1063 Set Similarity 

n个序列分别先放进集合里去重。在询问的时候，遍历A集合中每个数，判断下该数在B集合中是否存在，统计存在个数（分子），分母就是两个集合大小减去分子。

  1.  // 1063 Set Similarity
  2.  #include <set>
  3.  #include <map>
  4.  #include <cstdio>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  set <int> se[];
  10.  
  11.  int main() {
  12.      int n, k, m;
  13.      scanf("%d", &n);
  14.      for(int i = ; i <= n; i++) {
  15.          scanf("%d", &m);
  16.          while(m--) {
  17.              scanf("%d", &k);
  18.              se[i].insert(k);
  19.          }
  20.      }
  21.      scanf("%d", &k);
  22.      while(k--) {
  23.          int id1, id2;
  24.          double cnt = , total = ;
  25.          scanf("%d %d", &id1, &id2);
  26.          for(int i : se[id1]) {
  27.              if(se[id2].find(i) != se[id2].end()) cnt++;
  28.          }
  29.          total = se[id1].size() + se[id2].size() - cnt;
  30.          printf("%.1f%\n", cnt *  / total);
  31.      }
  32.      return ;
  33.  }

1064 Complete Binary Search Tree 

按照中序遍历建树。对应根节点依次填入排序好的数即可。

  1.  // 1064 Complete Binary Search Tree
  2.  #include <queue>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = ;
  9.  int a[N], level[N], n, cnt = ;
  10.  
  11.  void solve(int root) {
  12.      if(root > n) return ;
  13.      solve( * root);
  14.      level[root] = a[++cnt];
  15.      solve( * root + );
  16.  }
  17.  
  18.  int main() {
  19.      scanf("%d", &n);
  20.      for(int i = ; i <= n; i++) scanf("%d", &a[i]);
  21.      sort(a + , a +  + n);
  22.      solve();
  23.      for(int i = ; i <= n; i++) {
  24.          if(i != ) printf(" ");
  25.          printf("%d", level[i]);
  26.      }
  27.      return ;
  28.  }

1065 A+B and C (64bit)

对溢出的情况特判一下。

  1.  // 1065 A+B and C (64bit)
  2.  #include <iostream>
  3.  using namespace std;
  4.  
  5.  typedef long long ll;
  6.  
  7.  int main() {
  8.      int t;
  9.      scanf("%d", &t);
  10.      for(int i = ; i <= t; i++) {
  11.          ll a, b, c, sum = ;
  12.          scanf("%lld %lld %lld", &a, &b, &c);
  13.          sum = a + b;
  14.          if(a >  && b >  && sum <= ) {
  15.              printf("Case #%d: true\n", i);
  16.          } else if(a <  && b <  && sum >= ) {
  17.              printf("Case #%d: false\n", i);
  18.          } else if(sum > c) {
  19.              printf("Case #%d: true\n", i);
  20.          } else {
  21.              printf("Case #%d: false\n", i);
  22.          }
  23.      }
  24.      return ;
  25.  }

1066 Root of AVL Tree 

AVL模板题

  1.  #include <bits/stdc++.h>
  2.  using namespace std;
  3.  
  4.  typedef struct node;
  5.  typedef node * tree;
  6.  
  7.  struct node{
  8.      int v, heigh;
  9.      tree L,R;
  10.  };
  11.  
  12.  int getheigh(tree root){
  13.      if(root==NULL) return ;
  14.      return root->heigh;
  15.  }
  16.  
  17.  void updataheigh(tree root){
  18.      root->heigh=max(getheigh(root->L),getheigh(root->R))+;
  19.  }
  20.  
  21.  int getBalance(tree root){
  22.      return getheigh(root->L)-getheigh(root->R);
  23.  }
  24.  
  25.  void L(tree &root){
  26.      tree temp;
  27.      temp=root->R;
  28.      root->R=temp->L;
  29.      temp->L=root;
  30.      updataheigh(root);
  31.      updataheigh(temp);
  32.      root=temp;
  33.  }
  34.  
  35.  void R(tree &root){
  36.      tree temp;
  37.      temp=root->L;
  38.      root->L=temp->R;
  39.      temp->R=root;
  40.      updataheigh(root);
  41.      updataheigh(temp);
  42.      root=temp;
  43.  }
  44.  
  45.  void insertt(tree &root,int v){
  46.      if(root==NULL){
  47.          root=new node;
  48.          root->v=v;
  49.          root->heigh=;
  50.          root->L=root->R=NULL;
  51.          return;
  52.      }
  53.      if(v<root->v){
  54.          insertt(root->L,v);
  55.          updataheigh(root);
  56.          if(getBalance(root)==){
  57.              if(getBalance(root->L)==){
  58.                  R(root);
  59.              }
  60.              else if(getBalance(root->L)==-){
  61.                  L(root->L);
  62.                  R(root);
  63.              }
  64.          }
  65.      }
  66.      else{
  67.          insertt(root->R,v);
  68.          updataheigh(root);
  69.          if(getBalance(root)==-){
  70.              if(getBalance(root->R)==-){
  71.                  L(root);
  72.              }
  73.              else if(getBalance(root->R)==){
  74.                  R(root->R);
  75.                  L(root);
  76.              }
  77.          }
  78.      }
  79.  
  80.  }
  81.  
  82.  int main(){
  83.      int n;
  84.      scanf("%d",&n);
  85.      int x;
  86.      tree root;
  87.      root=NULL;
  88.      for(int i=;i<n;i++){
  89.          scanf("%d",&x);
  90.          insertt(root,x);
  91.      }
  92.      printf("%d\n",root->v);
  93.      return ;
  94.  }

1067 Sort with Swap(0, i)

看到表排序的知识，摘录下。

表排序思想：N个数字的排列由若干个独立的环组成

分成以下三种环：

1. 环内只有一个元素，即该数已经在正确的位置上，不需要交换。
2. 环内有 n 个元素，含 0，需要交换次数是：n-1 （每一个非 0 元素都要和 0 换一下）
3. 环内有 n 个元素，不含 0，需要交换的次数是：n+1（把 0 换入换出 + 其余 n-1 个非 0 元素和 0 换一下）

  1.  // 1067 Sort with Swap(0, i)
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  const int N = 1e5 + ;
  8.  bool vis[N];
  9.  int pos[N];
  10.  
  11.  int main() {
  12.      int n, u, ans = ;
  13.      scanf("%d", &n);
  14.      for(int i = ; i < n; i++) {
  15.          scanf("%d", &u);
  16.          pos[u] = i;
  17.      }
  18.      for(int i = ; i < n; i++) {
  19.          if(pos[i] != i && !vis[i]) {
  20.              int cnt = , j = i;
  21.              do {
  22.                  cnt++;
  23.                  vis[j] = true;
  24.                  if(j == ) cnt -= ;
  25.                  j = pos[j];
  26.              }while(j != i);
  27.              cnt++;
  28.              ans += cnt;
  29.          }
  30.      }
  31.      printf("%d\n", ans);
  32.      return ;
  33.  }

1068 Find More Coins

因题目特殊的输出要求，用01背包的时候需要先用大的数填，不然如果先用小的数填，小的数会影响后面的数，最后从小的往大的拿出去即可。

  1.  // 1068 Find More Coins
  2.  #include <vector>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e4 + ;
  9.  int n, m;
  10.  int a[N], dp[N];
  11.  vector <int> ans;
  12.  bool check[N][];
  13.  
  14.  int main() {
  15.      scanf("%d %d", &n, &m);
  16.      for(int i = ; i <= n; i++) scanf("%d", &a[i]);
  17.      sort(a + , a +  + n);
  18.      for(int i = n; i >= ; i--) {
  19.          for(int j = m; j >= a[i]; j--) {
  20.              if(dp[j - a[i]] + a[i] >= dp[j]) {
  21.                  dp[j] = dp[j - a[i]] + a[i];
  22.                  check[i][j] = ;
  23.              }
  24.          }
  25.      }
  26.      if(dp[m] != m) {
  27.          printf("No Solution\n");
  28.          return ;
  29.      }
  30.      int id = ;
  31.      while(m > ) {
  32.          if(check[id][m]) {
  33.              ans.push_back(a[id]);
  34.              m -= a[id];
  35.          }
  36.          id++;
  37.      }
  38.      for(int i = ; i < ans.size(); i++) {
  39.          if(i != ) printf(" ", ans[i]);
  40.          printf("%d", ans[i]);
  41.      }
  42.      return ;
  43.  }

1069 The Black Hole of Numbers

按题目模拟即可。

  1.  // 1069 The Black Hole of Numbers
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  int a[];
  9.  bool cmp(int a, int b) {return a > b;}
  10.  
  11.  int cal1(int n) {
  12.      memset(a, , sizeof(a));
  13.      int cnt = , res = ;
  14.      while(n) {
  15.          a[++cnt] = n % ;
  16.          n /= ;
  17.      }
  18.      sort(a + , a +  + );
  19.      for(int i = ; i >= ; i--) {
  20.          res = res *  + a[i];
  21.      }
  22.      return res;
  23.  }
  24.  
  25.  int cal2() {
  26.      int res = ;
  27.      sort(a + , a +  + );
  28.      for(int i = ; i <= ; i++) {
  29.          res = res *  + a[i];
  30.      }
  31.      return res;
  32.  
  33.  }
  34.  
  35.  int main() {
  36.      int n;
  37.      scanf("%d", &n);
  38.      while() {
  39.          int p1 = cal1(n);
  40.          int p2 = cal2();
  41.          n = p1 - p2;
  42.          printf("%04d - %04d = %04d\n", p1, p2, n);
  43.          if(n ==  || n == ) return ;
  44.      }
  45.      return ;
  46.  }

1070 Mooncake

按每吨多少钱从大到小排序下，再取即可。

  1.  // 1070 Mooncake
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  const int N = ;
  8.  
  9.  struct node{
  10.      double storage, price;
  11.  }moon[N];
  12.  
  13.  bool cmp(node x, node y) {
  14.      return x.price * y.storage  > y.price * x.storage;
  15.  }
  16.  
  17.  int main() {
  18.      int n;
  19.      double d, ans = ;
  20.      scanf("%d %lf", &n, &d);
  21.      for(int i = ; i <= n; i++) scanf("%lf", &moon[i].storage);
  22.      for(int i = ; i <= n; i++) scanf("%lf", &moon[i].price);
  23.      sort(moon + , moon +  + n, cmp);
  24.      for(int i = ; i <= n; i++) {
  25.          if(d >= moon[i].storage) {
  26.              ans += moon[i].price;
  27.              d -= moon[i].storage;
  28.          } else {
  29.              ans += moon[i].price * (1.0 * d / moon[i].storage);
  30.              break;
  31.          }
  32.      }
  33.      printf("%.2f\n", ans);
  34.      return ;
  35.  }

1071 Speech Patterns

用map存下。更新答案即可。

  1.  // 1071 Speech Patterns
  2.  #include <map>
  3.  #include <string>
  4.  #include <iostream>
  5.  using namespace std;
  6.  
  7.  string in, tmp, ans1;
  8.  map<string, int> mp;
  9.  map<string, int>::iterator it;
  10.  
  11.  int main(){
  12.      int ans2 = ;
  13.      getline(cin, in);
  14.      for(int i =  ; i < in.size(); i++) {
  15.          if(isalnum(in[i])) {
  16.              tmp += tolower(in[i]);
  17.              if(i == in.size() - ) mp[tmp]++;
  18.          } else {
  19.              if(tmp.size()) mp[tmp]++;
  20.              tmp.clear();
  21.          }
  22.      }
  23.      for(it = mp.begin(); it!=mp.end(); it++){
  24.          if(ans2 < it->second){
  25.              ans2 = it -> second;
  26.              ans1 = it -> first;
  27.          }else if(ans2 == it -> second && ans1 > it -> first){
  28.              ans1 = it -> first;
  29.          }
  30.      }
  31.      cout << ans1 << " " << ans2 << endl;
  32.  }

1072 Gas Station

以加油站为起点，跑最短路即可。

  1.  //1072 Gas Station
  2.  #include <queue>
  3.  #include <cstdio>
  4.  #include <cstring>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N=1e3+;
  10.  const int INF=0x3f3f3f3f;;
  11.  int n,m,k,ds,mp[N][N],dis[N],vis[N];
  12.  
  13.  int toint(char s[]){
  14.      int sum=;
  15.      int len=strlen(s);
  16.      for(int i=;i<len;i++){
  17.          if(s[i]=='G') continue;
  18.          sum=sum*+s[i]-'';
  19.      }
  20.      if(s[]=='G') return sum+n;
  21.      return sum;
  22.  }
  23.  
  24.  void Dijkstra(int v){
  25.      fill(vis,vis+N,false);
  26.      fill(dis,dis+N,INF);
  27.      for(int i=;i<=n+m;i++) dis[i]=mp[v][i];
  28.      dis[v]=;
  29.      for(int i=;i<n+m;i++){
  30.          int u=-,minn=INF;
  31.          for(int j=;j<=n+m;j++){
  32.              if(!vis[j]&&minn>dis[j]){
  33.                  u=j;minn=dis[j];
  34.              }
  35.          }
  36.          vis[u]=true;
  37.          for(int j=;j<=n+m;j++){
  38.              if(!vis[j]&&dis[j]>mp[u][j]+dis[u]){
  39.                  dis[j]=mp[u][j]+dis[u];
  40.              }
  41.          }
  42.      }
  43.  }
  44.  int main(){
  45.      int num,a,b;
  46.      char s1[],s2[];
  47.      scanf("%d %d %d %d",&n,&m,&k,&ds);
  48.      fill(mp[],mp[]+N*N,INF);
  49.      for(int i=;i<k;i++){
  50.          scanf("%s %s %d",s1,s2,&num);
  51.          a=toint(s1),b=toint(s2);
  52.          mp[a][b]=mp[b][a]=num;
  53.      }
  54.      int id=-;
  55.      double maxDis=-,minAvg=INF;
  56.      for(int i=n+;i<=n+m;i++){
  57.          Dijkstra(i);
  58.          double avg=;
  59.          double Min=INF;
  60.          bool flag=false;
  61.          for(int j=;j<=n;j++){
  62.              if(dis[j]>ds){
  63.                  flag=true;
  64.                  break;
  65.              }
  66.              if(Min>dis[j]){
  67.                  Min=dis[j];
  68.              }
  69.              avg+=1.0*dis[j]/n;
  70.          }
  71.          if(flag) continue;
  72.          if(maxDis<Min){
  73.              maxDis=Min;
  74.              id=i;
  75.              minAvg=avg;
  76.          }
  77.          else if(maxDis==Min&&minAvg>avg){
  78.              id=i;
  79.              minAvg=avg;
  80.          }
  81.  
  82.      }
  83.      if(id==-) printf("No Solution\n");
  84.      else{
  85.          printf("G%d\n",id-n);
  86.          printf("%.1f %.1f\n",maxDis,minAvg);
  87.      }
  88.      return ;
  89.  }

1073 Scientific Notation

按题意模拟即可。

  1.  // 1073 Scientific Notation
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e4 + ;
  9.  char s[N];
  10.  
  11.  int main() {
  12.      int n, i = ;
  13.      scanf("%s", s);
  14.      n = strlen(s);
  15.      while(s[i] != 'E') {
  16.          i++;
  17.      }
  18.      i++;
  19.      if(s[] != '+') printf("%c", s[]);
  20.      int m = ;
  21.      for(int j = i + ; j < n; j++) m =  * m + (s[j] - '');
  22.      if(s[i] == '+') {
  23.          m+=;
  24.          for(int j = ; j < i - ; j++) {
  25.              m--;
  26.              if(m == -) printf(".");
  27.              if(s[j] != '.') printf("%c", s[j]);
  28.          }
  29.          while(m > ) {
  30.              printf("");
  31.              m--;
  32.          }
  33.      } else {
  34.          m--;
  35.          printf("0.");
  36.          while(m > ) {
  37.              printf("");
  38.              m--;
  39.          }
  40.          for(int j = ; j < i - ; j++) {
  41.              if(s[j] != '.') printf("%c", s[j]);
  42.          }
  43.      }
  44.      return ;
  45.  }

1074 Reversing Linked List

每k个长度翻转一次。$j$位置对应交换的位置为$(2 * i + k - j - 1)$，这个只使用前半和后半交换，注意特判跳出。

  1.  //1074 Reversing Linked List
  2.  #include <cstdio>
  3.  #include <vector>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e5 + ;
  9.  int nxt[N], val[N], res[N];
  10.  
  11.  int main() {
  12.      int root, n, k, cnt = ;
  13.      scanf("%d%d%d", &root, &n, &k);
  14.      for(int i = ; i <= n; i++) {
  15.          int u, v, w;
  16.          scanf("%d%d%d", &u, &v, &w);
  17.          val[u] = v; nxt[u] = w;
  18.      }
  19.      while(root != -) {
  20.          res[cnt++] = root;
  21.          root = nxt[root];
  22.      }
  23.      for(int i = ; i + k <= cnt; i += k) {
  24.          for(int j = i; j < (i + k); j++) {
  25.              if(j >= ( * i + k - j - )) break;
  26.              swap(res[( * i + k - j - )], res[j]);
  27.          }
  28.      }
  29.      for(int i = ; i < cnt - ; i++) {
  30.          printf("%05d %d %05d\n", res[i], val[res[i]], res[i + ]);
  31.      }
  32.      printf("%05d %d -1\n", res[cnt - ], val[res[cnt - ]]);
  33.      return ;
  34.  }

1075 PAT Judge

结构体排序，注意一下区别提交得0分和提交没有通过编译。

  1.  // 1075 PAT Judge
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e5 + ;
  9.  int v[];
  10.  int n, k, m;
  11.  struct node{
  12.      int q[], id, sum, ok, perfect;
  13.  }p[N];
  14.  
  15.  void init() {
  16.      for(int i = ; i <= n; i++)
  17.      for(int j = ; j <= k; j++)
  18.      p[i].q[j] = -, p[i].id = i;
  19.  }
  20.  
  21.  bool cmp(node x, node y) {
  22.      if(x.sum == y.sum) {
  23.          if(x.perfect == y.perfect) return x.id < y.id;
  24.          return x.perfect > y.perfect;
  25.      }
  26.      return x.sum > y.sum;
  27.  }
  28.  
  29.  int main() {
  30.      int r = ;
  31.      scanf("%d %d %d", &n, &k, &m);
  32.      for(int i = ; i <= k; i++) scanf("%d", &v[i]);
  33.      init();
  34.      for(int i = ; i <= m; i++) {
  35.          int uid, pid, val;
  36.          scanf("%d %d", &uid, &pid);
  37.          scanf("%d", &val);
  38.          p[uid].q[pid] = max(p[uid].q[pid], val);
  39.      }
  40.      for(int i = ; i <= n; i++) {
  41.          int cnt = , perfect = ;
  42.          bool ok = ;
  43.          for(int j = ; j <= k; j++) {
  44.              if(p[i].q[j] != - && p[i].q[j] != -) cnt += p[i].q[j], ok = ;
  45.              if(p[i].q[j] == v[j]) perfect++;
  46.          }
  47.          if(ok) p[i].ok = ;
  48.          else p[i].ok = ;
  49.          p[i].sum = cnt;
  50.          p[i].perfect = perfect;
  51.      }
  52.      sort(p + , p +  + n, cmp);
  53.      for(int i = ; i <= n; i++) {
  54.          if(p[i].ok == ) continue;
  55.          if(p[i].sum != p[i - ].sum) r = i;
  56.          printf("%d %05d %d", r, p[i].id, p[i].sum);
  57.          for(int j = ; j <= k; j++) {
  58.              if(p[i].q[j] == -) printf(" -");
  59.              else if(p[i].q[j] == -) printf("");
  60.              else printf(" %d",p[i].q[j]);
  61.          }
  62.          printf("\n");
  63.      }
  64.      return ;
  65.  }

1076 Forwards on Weibo

DFS或BFS，DFS的时候注意剪枝。

DFS版本

  1.  //1076 Forwards on Weibo
  2.  #include <vector>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = ;
  9.  vector <int> E[N];
  10.  bool vis[N];
  11.  int layer[N];
  12.  int n, m, k, u, sz, ans;
  13.  
  14.  void dfs(int u, int l) {
  15.      vis[u] = ;
  16.      layer[u] = l;
  17.      if(l <= k) {
  18.          for(int i = ; i < E[u].size(); i++) {
  19.              int v = E[u][i];
  20.              if(!vis[v] || layer[v] > l + ) {
  21.                  dfs(v, l + );
  22.              }
  23.          }
  24.      }
  25.  }
  26.  
  27.  int main() {
  28.      scanf("%d %d", &n, &k);
  29.      for(int i = ; i <= n; i++) {
  30.          scanf("%d", &sz);
  31.          while(sz) {
  32.              sz--;
  33.              scanf("%d", &u);
  34.              E[u].push_back(i);
  35.          }
  36.      }
  37.      scanf("%d", &m);
  38.      while(m--) {
  39.          for(int i = ; i <= n; i++) vis[i] = , layer[i] = ;
  40.          scanf("%d", &u);
  41.          ans = ;
  42.          dfs(u, );
  43.          for(int i = ; i <= n; i++) if(layer[i] > ) ans++;
  44.          printf("%d\n", ans - );
  45.      }
  46.      return ;
  47.  }

BFS版本

  1.  //1076 Forwards on Weibo
  2.  #include <queue>
  3.  #include <vector>
  4.  #include <cstdio>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N = ;
  10.  vector <int> E[N];
  11.  bool vis[N];
  12.  int n, m, k, u, sz;
  13.  
  14.  void bfs() {
  15.      int ans = ;
  16.      for(int i = ; i <= n; i++) vis[i] = ;
  17.      queue < pair<int,int> > Q;
  18.      Q.push(make_pair(u, ));
  19.      vis[u] = ;
  20.      while(!Q.empty()) {
  21.          int cur = Q.front().first;
  22.          int layer = Q.front().second;
  23.          Q.pop();
  24.          if(layer >= k) continue;
  25.          for(int i = ; i < E[cur].size(); i++) {
  26.              int v = E[cur][i];
  27.              if(!vis[v]) {
  28.                  ans++;
  29.                  vis[v] = ;
  30.                  Q.push(make_pair(v, layer + ));
  31.              }
  32.          }
  33.      }
  34.      printf("%d\n", ans);
  35.  }
  36.  
  37.  int main() {
  38.      scanf("%d %d", &n, &k);
  39.      for(int i = ; i <= n; i++) {
  40.          scanf("%d", &sz);
  41.          while(sz) {
  42.              sz--;
  43.              scanf("%d", &u);
  44.              E[u].push_back(i);
  45.          }
  46.      }
  47.      scanf("%d", &m);
  48.      while(m--) {
  49.          scanf("%d", &u);
  50.          bfs();
  51.      }
  52.      return ;
  53.  }

1077 Kuchiguse

模拟

  1.  //1077 Kuchiguse
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  int id[];
  8.  char s[][];
  9.  
  10.  int main() {
  11.      int n, cnt = ;
  12.      scanf("%d", &n);
  13.      getchar();
  14.      for(int i = ; i <= n; i++) {
  15.          char c;
  16.          while(scanf("%c", &c) != EOF && c != '\n') {
  17.              s[i][id[i]++] = c;
  18.          }
  19.      }
  20.      while() {
  21.          for(int i = ; i <= n; i++) {
  22.              if( ((id[i] - cnt) < )  || (s[i][id[i] - cnt] != s[][id[] - cnt]) ) {
  23.                  cnt--;
  24.                  if(cnt == ) printf("nai");
  25.                  else {
  26.                      for(int i = id[] - cnt; i < id[] ; i++) {
  27.                          printf("%c", s[][i]);
  28.                      }
  29.                  }
  30.                  return ;
  31.              }
  32.          }
  33.          cnt++;
  34.      }
  35.      return ;
  36.  }

1078 Hashing

Hash二次探测。0^2，1^2，2^2...。

  1.  //1078 Hashing
  2.  #include <iostream>
  3.  using namespace std;
  4.  
  5.  const int N = 1e4 + ;
  6.  bool vis[N];
  7.  
  8.  bool check(int x) {
  9.      if(x < ) return false;
  10.      for(int i = ; i * i <= x; i++) {
  11.          if(x % i == ) return false;
  12.      }
  13.      return true;
  14.  }
  15.  
  16.  int main() {
  17.      int n, m, p;
  18.      cin >> m >> n;
  19.      while(!check(m)) {m++;}
  20.      for(int i = ; i <= n; i++) {
  21.          bool f = ;
  22.          cin >> p;
  23.          if(i != ) printf(" ");
  24.          for(int i = ; i < m; i++) {
  25.              int tmp = (p + i * i) % m;
  26.              if(!vis[tmp]) {
  27.                  f = ;
  28.                  printf("%d", tmp);
  29.                  vis[tmp] = ;
  30.                  break;
  31.              }
  32.          }
  33.          if(!f) printf("-");
  34.      }
  35.      return ;
  36.  }

1079 Total Sales of Supply Chain

从根跑一遍DFS到叶子节点。统计每个叶子节点的能够产生的销售额。

  1.  //1079 Total Sales of Supply Chain
  2.  #include <vector>
  3.  #include <cstdio>
  4.  using namespace std;
  5.  
  6.  const int N = 1e5 + ;
  7.  int cnt[N];
  8.  bool vis[N];
  9.  double p, r;
  10.  double price[N];
  11.  vector <int> E[N];
  12.  
  13.  void dfs(int u, double val) {
  14.      vis[u] = ; price[u] = val;
  15.      for(int i = ; i < E[u].size(); i++) {
  16.          int v = E[u][i];
  17.          if(!vis[v]) dfs(v, val * ( + r / ));
  18.      }
  19.  }
  20.  
  21.  int main() {
  22.      int n, k, u;
  23.      scanf("%d %lf %lf", &n, &p, &r);
  24.      for(int i = ; i < n; i++) {
  25.          scanf("%d", &k);
  26.          if(k == ){
  27.               scanf("%d", &cnt[i]);
  28.          } else {
  29.              while(k--) {
  30.                  scanf("%d", &u);
  31.                  E[i].push_back(u);
  32.              }
  33.          }
  34.      }
  35.      double ans = ;
  36.      dfs(, p);
  37.      for(int i = ; i < n; i++) {
  38.          ans += price[i] * cnt[i];
  39.      }
  40.      printf("%.1f\n", ans);
  41.      return ;
  42.  }

1080 Graduate Admission

按照题意模拟，用set从小到大排序特性会方便很多。

  1.  // 1080 Graduate Admission
  2.  #include <set>
  3.  #include <cstdio>
  4.  #include <vector>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N = 4e4 + ;
  10.  struct Stu{
  11.    int Id, Ge, Gi, Gh, Rank;
  12.    int choices[];
  13.  };
  14.  
  15.  struct Sch{
  16.    int maxn, lastRank;
  17.    set<int> admit;
  18.  };
  19.  
  20.  bool cmp(Stu x, Stu y){
  21.      if(x.Gh == y.Gh) return x.Ge > y.Ge;
  22.      return x.Gh > y.Gh;
  23.  }
  24.  
  25.  set<int>::iterator it;
  26.  
  27.  int main(){
  28.      int n, m, k;
  29.      scanf("%d%d%d", &n, &m, &k);
  30.      vector<Sch> sch(m);
  31.      vector<Stu> stu(n);
  32.      for(int i = ; i < m; i++) scanf("%d", &sch[i].maxn);
  33.      for(int i = ; i < n; i++){
  34.          scanf("%d%d", &stu[i].Ge, &stu[i].Gi);
  35.          stu[i].Id=i; stu[i].Gh = stu[i].Ge + stu[i].Gi;
  36.          for(int j = ; j < k; j++) scanf("%d", &stu[i].choices[j]);
  37.      }
  38.      sort(stu.begin(), stu.end(), cmp);
  39.      int Rank = ;
  40.      for(int i = ; i < n; i++){
  41.          if(i== || (stu[i].Gh == stu[i-].Gh && stu[i].Gi == stu[i-].Gi)) stu[i].Rank = Rank;
  42.          else{
  43.              Rank = i + ;
  44.              stu[i].Rank = Rank;
  45.          }
  46.      }
  47.      for(int i = ; i < n; i++){
  48.          bool f=;
  49.          for(int j = ; j < k; j++){
  50.              int schid = stu[i].choices[j];
  51.              if(sch[schid].maxn > ){
  52.                  sch[schid].admit.insert(stu[i].Id);
  53.                  sch[schid].maxn--;
  54.                  sch[schid].lastRank = stu[i].Rank;
  55.                  f = true;
  56.              }else if(sch[schid].lastRank == stu[i].Rank) {
  57.                  sch[schid].admit.insert(stu[i].Id);
  58.                  f = true;
  59.              }
  60.              if(f) break;
  61.          }
  62.      }
  63.      for(int i = ; i < m; i++){
  64.          for(it = sch[i].admit.begin(); it!=sch[i].admit.end(); it++){
  65.            if(it == sch[i].admit.begin()) printf("%d", *it);
  66.            else printf(" %d", *it);
  67.          }
  68.          printf("\n");
  69.      }
  70.      return ;
  71.  }

1081 Rational Sum

通分，注意最后若答案为0的时的格式输出。

  1.  // 1081 Rational Sum
  2.  #include <cstdio>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  typedef long long ll;
  7.  ll a[], b[];
  8.  
  9.  int main() {
  10.      ll n, nr = , dm = ;
  11.      scanf("%lld", &n);
  12.      for(ll i = ; i <= n; i++) {
  13.          scanf("%lld/%lld", &a[i], &b[i]);
  14.          dm = b[i] / __gcd(b[i], dm) * dm;
  15.      }
  16.      for(ll i = ; i<= n; i++) {
  17.          nr = nr + (dm / b[i]) * a[i];
  18.      }
  19.      ll g = __gcd(nr, dm);
  20.      if(g != ) nr /= g, dm /= g;
  21.      if(nr / dm) {
  22.          printf("%lld", nr / dm);
  23.          nr = nr - (nr / dm) * dm;
  24.          if(nr != ) printf(" %lld/%lld", nr, dm);
  25.          return ;
  26.      }
  27.      nr = nr - (nr / dm) * dm;
  28.      if(nr != ) printf("%lld/%lld", nr, dm);
  29.      else printf("");
  30.      return ;
  31.  }

1082 Read Number in Chinese

模拟，注意细节即可。

  1.  // 1082 Read Number in Chinese
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  string a[] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
  8.  string b[] = {"","Shi","Bai","Qian","Wan","Shi","Bai","Qian","Yi"};
  9.  
  10.  int main() {
  11.      int n, f = ;
  12.      string s;
  13.      cin >> s;
  14.      if(s[] == '-') {
  15.          cout << "Fu" << " ";
  16.          s = s.substr();
  17.      }
  18.      n = s.size();
  19.      reverse(s.begin(), s.end());
  20.      for(int i = n - ; i >= ; i--) {
  21.          if(i == n - ) {
  22.              cout << a[s[i] - ''] << " " << b[i];
  23.          } else {
  24.              if(s[i] != '') {
  25.                  if(f) {
  26.                      cout << " " << a[];
  27.                      f = ;
  28.                  }
  29.                  cout << " " << a[s[i] - ''] << " " << b[i];
  30.              } else {
  31.                  f = ;
  32.                  if(i == ) cout << " " << b[];
  33.              }
  34.          }
  35.      }
  36.      if(f && s[] != '') {
  37.          cout << " " << a[];
  38.      }
  39.      if(s.size() == ) {
  40.          cout << a[s[] - ''];
  41.      }
  42.      else if(s.size() >  && s[] != ''){
  43.          cout << " " << a[s[] - ''];
  44.      }
  45.      return ;
  46.  }

1083 List Grades

结构体排序

  1.  // 1083 List Grades
  2.  #include <cstdio>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  const int N = 2e5 + ;
  7.  struct node {
  8.      char name[], id[];
  9.      int grade;
  10.  }recoder[N];
  11.  
  12.  bool cmp(node x, node y) {
  13.      return x.grade > y.grade;
  14.  }
  15.  
  16.  int main() {
  17.      bool f = ;
  18.      int n;
  19.      scanf("%d", &n);
  20.      for(int i = ; i <= n; i++) {
  21.          scanf("%s %s %d", recoder[i].name, recoder[i].id, &recoder[i].grade);
  22.      }
  23.      int g1, g2;
  24.      scanf("%d%d", &g1, &g2);
  25.      sort(recoder + , recoder +  + n, cmp);
  26.      for(int i = ; i <= n; i++) {
  27.          if(recoder[i].grade <= g2 && recoder[i].grade >= g1) {
  28.              printf("%s %s\n", recoder[i].name, recoder[i].id);
  29.              f = ;
  30.          }
  31.      }
  32.      if(!f) printf("NONE\n");
  33.      return ;
  34.  }

map标记一下。

  1.  // 1084 Broken Keyboard
  2.  #include <map>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  map <char, int> m;
  9.  
  10.  int main() {
  11.      char c;
  12.      string s1, s2;
  13.      cin >> s1 >> s2;
  14.      for(int i = ; i < s2.size(); i++) {
  15.          m[s2[i]] = ;
  16.          if(s2[i] >= 'a' && s2[i] <= 'z') {
  17.              c = s2[i] - ;
  18.              m[c] = ;
  19.          }
  20.          else if(s2[i] >= 'A' && s2[i] <= 'Z') {
  21.              c = s2[i] + ;
  22.              m[c] = ;
  23.          }
  24.      }
  25.      for(int i = ; i < s1.size(); i++) {
  26.          if(!m[s1[i]]) {
  27.              m[s1[i]] = ;
  28.              if(s1[i] >= 'a' && s1[i] <= 'z') {
  29.                  c = s1[i] - ;
  30.                  m[c] = ;
  31.                  cout << c;
  32.              }
  33.              else if(s1[i] >= 'A' && s1[i] <= 'Z') {
  34.                  c = s1[i] + ;
  35.                  m[c] = ;
  36.                  cout << s1[i];
  37.              } else {
  38.                  cout << s1[i];
  39.              }
  40.          }
  41.      }
  42.      return ;
  43.  }

1085 Perfect Sequence

二分搜索

  1.  // 1085 Perfect Sequence
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  typedef long long ll;
  8.  const int N = 1e5 + ;
  9.  ll a[N];
  10.  
  11.  int main() {
  12.      int ans = ;
  13.      ll n, p;
  14.      scanf("%lld %lld", &n, &p);
  15.      for(int i = ; i < n; i++) scanf("%lld", &a[i]);
  16.      sort(a, a + n);
  17.      for(int i = ; i < n; i++) {
  18.          int pos1 = lower_bound(a, a + n, a[i]) - a;
  19.          int pos2 = lower_bound(a, a + n, a[i] * p + ) - a;
  20.          pos2--;
  21.          ans = max(ans, pos2 - pos1 + );
  22.      }
  23.      printf("%d\n", ans);
  24.      return ;
  25.  }

1086 Tree Traversals Again

由题目可以得知push的时候为往下找，分情况讨论存左孩子还是右孩子，pop的时候表示往右走，下一次push存右孩子，最后后序遍历一下即可。

  1.  // 1086 Tree Traversals Again
  2.  #include <stack>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  char s[];
  9.  struct node{
  10.      int lch, rch;
  11.  }tree[];
  12.  stack <int> sta;
  13.  int n, u, fa, root = -, child = ;
  14.  
  15.  void postorder(int u) {
  16.      if(tree[u].lch != ) postorder(tree[u].lch);
  17.      if(tree[u].rch != ) postorder(tree[u].rch);
  18.      if(u == root) printf("%d", u);
  19.      else printf("%d ", u);
  20.  }
  21.  
  22.  int main() {
  23.      scanf("%d", &n);
  24.      n *= ;
  25.      while(n--) {
  26.          scanf("%s", s);
  27.          if(s[] == 'u') {
  28.              scanf("%d", &u);
  29.              sta.push(u);
  30.              if(root == -) {
  31.                  root = u;
  32.              } else {
  33.                  if(child == ) tree[fa].lch = u;
  34.                  else tree[fa].rch = u;
  35.              }
  36.              fa = u;
  37.              child = ;
  38.          } else {
  39.              if(sta.size())
  40.              fa = sta.top();
  41.              sta.pop();
  42.              child = ;
  43.          }
  44.      }
  45.      postorder(root);
  46.      return ;
  47.  }

1087 All Roads Lead to Rome

先Floyd得到最短路，再DFS搜，搜的过程中更新答案。

  1.  // 1087 All Roads Lead to Rome
  2.  #include <bits/stdc++.h>
  3.  using namespace std;
  4.  
  5.  const int N = ;
  6.  int n, m, cost, cnt = , maxhappy = ;
  7.  int happy[N], d[N][N];
  8.  string s[N], s1, s2;
  9.  map<string, int> mp;
  10.  vector<int> E[N], res, ans;
  11.  bool vis[N];
  12.  
  13.  void dfs(int u, int en, int co) {
  14.      if(u == en && co == cost) {
  15.          int sum = ;
  16.          for(int i = ; i < res.size(); i++) {
  17.              sum += happy[res[i]];
  18.          }
  19.          if(sum > maxhappy) {
  20.              maxhappy = sum;
  21.              ans.clear();
  22.              for(int i = ; i < res.size(); i++) ans.push_back(res[i]);
  23.          }
  24.          else if(sum == maxhappy && res.size() < ans.size()) {
  25.              ans.clear();
  26.              for(int i = ; i < res.size(); i++) ans.push_back(res[i]);
  27.          }
  28.          cnt++;
  29.          return ;
  30.      }
  31.      for(int k = ; k < E[u].size(); k++) {
  32.          int v = E[u][k];
  33.          if(vis[v]) continue;
  34.          res.push_back(v);
  35.          vis[v] = ;
  36.          dfs(v, en, co + d[u][v]);
  37.          vis[v] = ;
  38.          res.pop_back();
  39.      }
  40.  }
  41.  
  42.  int main() {
  43.      cin >> n >> m >> s[];
  44.      mp[s[]] = ;
  45.      memset(d, 0x3f, sizeof(d));
  46.      for(int i = ; i <= n; i++) d[i][i] = ;
  47.      for(int i = ; i <= n; i++) {
  48.          cin >> s[i] >> happy[i];
  49.          mp[s[i]] = i;
  50.      }
  51.      for(int i = ; i <= m; i++) {
  52.          cin >> s1 >> s2 >> cost;
  53.          d[mp[s1]][mp[s2]] = cost;
  54.          d[mp[s2]][mp[s1]] = cost;
  55.          E[mp[s2]].push_back(mp[s1]);
  56.          E[mp[s1]].push_back(mp[s2]);
  57.      }
  58.  
  59.      for(int k = ; k <= n; k++)
  60.      for(int i = ; i <= n; i++)
  61.      for(int j = ; j <= n; j++)
  62.      d[i][j] = min(d[i][j] ,d[i][k] + d[k][j]);
  63.  
  64.      cost = d[mp[s[]]][mp["ROM"]];
  65.      vis[mp[s[]]] = ;
  66.      dfs(mp[s[]], mp["ROM"], );
  67.  
  68.      cout << cnt << " " << cost << " " << maxhappy << " " << maxhappy / ans.size() << endl;
  69.      cout << s[];
  70.      for(int i = ; i < ans.size(); i++) cout << "->" << s[ans[i]];
  71.      return ;
  72.  }

1088 Rational Arithmetic

注意相乘会爆long long，判断负可以分别判断x 和 y

  1.  // 1088 Rational Arithmetic
  2.  #include <bits/stdc++.h>
  3.  using namespace std;
  4.  
  5.  typedef long long ll;
  6.  ll gcd(ll a, ll b) {
  7.      return b ==  ? a : gcd(b, a % b);
  8.  }
  9.  
  10.  void func(ll x, ll y) {
  11.      bool f = ;
  12.      if(x * y == ) {
  13.          if(y == ) printf("Inf");
  14.          else if(x == ) printf("");
  15.          return ;
  16.      }
  17.      if((x >  && y < ) || (x <  && y > )) f = ;
  18.      x = abs(x); y = abs(y);
  19.      ll val = x / y;
  20.      ll res = x - val * y;
  21.      if(f) printf("(-");
  22.      if(val != ) printf("%lld", val);
  23.      if(res == ) {
  24.          if(f) printf(")");
  25.          return ;
  26.      }
  27.      if(val != ) printf(" ");
  28.      ll g = gcd(res, y);
  29.      res /= g; y /= g;
  30.      printf("%lld/%lld", res, y);
  31.      if(f) printf(")");
  32.  }
  33.  
  34.  int main() {
  35.      ll a, b, c, d;
  36.      scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
  37.      func(a, b);printf(" + ");func(c, d);printf(" = ");func(a * d + b * c, b * d);printf("\n");
  38.      func(a, b);printf(" - ");func(c, d);printf(" = ");func(a * d - b * c, b * d);printf("\n");
  39.      func(a, b);printf(" * ");func(c, d);printf(" = ");func(a * c, b * d);printf("\n");
  40.      func(a, b);printf(" / ");func(c, d);printf(" = ");func(a * d, b * c);
  41.      return ;
  42.  }

1089 Insert or Merge

先判断是否为插入排序，不是的话，跑一遍归并，直到a数组和b数组一样，再进行一次归并操作。

  1.  // 1089 Insert or Merge
  2.  #include <iostream>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  const int N = ;
  7.  int a[N], b[N];
  8.  
  9.  int main() {
  10.      int n, i, j;
  11.      cin >> n;
  12.      for (i = ; i < n; i++) cin >> a[i];
  13.      for (i = ; i < n; i++) cin >> b[i];
  14.      for (i = ; i < n -  && b[i] <= b[i + ]; i++);
  15.      for (j = i + ; a[j] == b[j] && j < n; j++);
  16.      if (j == n) {
  17.          cout << "Insertion Sort" << endl;
  18.          sort(a, a + i + );
  19.      } else {
  20.          cout << "Merge Sort" << endl;
  21.          int k = , f = ;
  22.          while(f) {
  23.              f = ;
  24.              for (i = ; i < n; i++) {
  25.                  if (a[i] != b[i]) f = ;
  26.              }
  27.              k = k * ;
  28.              for (i = ; i < n / k; i++)
  29.              sort(a + i * k, a + (i + ) * k);
  30.              sort(a + n / k * k, a + n);
  31.          }
  32.      }
  33.      for(j = ; j < n; j++) {
  34.          if (j != )  cout << " ";
  35.          cout << a[j];
  36.      }
  37.      return ;
  38.  }

1090 Highest Price in Supply Chain

DFS跑一遍树即可。

  1.  // 1090 Highest Price in Supply Chain
  2.  #include <cstdio>
  3.  #include <vector>
  4.  using namespace std;
  5.  
  6.  const int N = 1e5 + ;
  7.  int n, u, cnt, root;
  8.  double p, r, maxp = , sellp[N];
  9.  vector <int> E[N];
  10.  
  11.  void dfs(int a, double price) {
  12.      sellp[a] = price;
  13.      maxp = max(maxp, price);
  14.      for(int i = ; i < E[a].size(); i++) {
  15.          int b = E[a][i];
  16.          dfs(b, price * (1.0 + r / ));
  17.      }
  18.  }
  19.  
  20.  int main() {
  21.      scanf("%d %lf %lf", &n, &p, &r);
  22.      for(int i = ; i < n; i++) {
  23.          scanf("%d", &u);
  24.          if(u == -) root = i;
  25.          else E[u].push_back(i);
  26.      }
  27.      dfs(root, p);
  28.      for(int i = ; i < n; i++) {
  29.          if(sellp[i] == maxp) cnt++;
  30.      }
  31.      printf("%.2f %d\n", maxp, cnt);
  32.      return ;
  33.  }

1091 Acute Stroke

3维BFS裸题。

  1.  // 1091 Acute Stroke
  2.  #include <queue>
  3.  #include <cstdio>
  4.  using namespace std;
  5.  
  6.  int n, m, l, t, ans = ;
  7.  int mp[][][];
  8.  bool vis[][][];
  9.  struct node{
  10.      int z, x, y;
  11.  };
  12.  int dx[] = {, , , , -, };
  13.  int dy[] = {, , -, , , };
  14.  int dz[] = {-, , , , , };
  15.  
  16.  bool check(int z, int x, int y) {
  17.      if(z >=  && z <= l && x >=  && x <= n && y >=  && y <= m) return true;
  18.      return false;
  19.  }
  20.  
  21.  int bfs(int z, int x, int y) {
  22.      queue <node> q;
  23.      node tmp;
  24.      int res = ;
  25.      q.push(node{z, x, y});
  26.      vis[z][x][y] = ;
  27.      while(!q.empty()) {
  28.          res++;
  29.          tmp = q.front();
  30.          q.pop();
  31.          for(int i = ; i < ; i++) {
  32.              int nz = tmp.z + dz[i];
  33.              int nx = tmp.x + dx[i];
  34.              int ny = tmp.y + dy[i];
  35.              if(check(nz, nx, ny) && !vis[nz][nx][ny] && mp[nz][nx][ny]) {
  36.                  vis[nz][nx][ny] = ;
  37.                  q.push(node{nz, nx, ny});
  38.              }
  39.          }
  40.  
  41.      }
  42.      if(res >= t) return res;
  43.      else return ;
  44.  }
  45.  
  46.  int main() {
  47.      scanf("%d%d%d%d", &n, &m, &l, &t);
  48.      for(int z = ; z <= l; z++)
  49.      for(int x = ; x <= n; x++)
  50.      for(int y = ; y <= m; y++)
  51.      scanf("%d", &mp[z][x][y]);
  52.  
  53.      for(int z = ; z <= l; z++)
  54.      for(int x = ; x <= n; x++)
  55.      for(int y = ; y <= m; y++)
  56.      if(mp[z][x][y] && !vis[z][x][y]) ans += bfs(z, x, y);
  57.  
  58.      printf("%d\n", ans);
  59.      return ;
  60.  }

1092 To Buy or Not to Buy

用数组标记一下即可。

  1.  // 1092 To Buy or Not to Buy
  2.  #include <iostream>
  3.  using namespace std;
  4.  
  5.  int a[];
  6.  string s1, s2;
  7.  
  8.  int main(){
  9.      cin >> s1 >> s2;
  10.      int c1 = , c2 = ;
  11.      for(int i = ; i < s1.size(); i++) a[s1[i]]++;
  12.      for(int i = ; i < s2.size(); i++) a[s2[i]]--;
  13.      for(int i = ; i < ; i++) {
  14.          if(a[i] < ) c1+=a[i];
  15.          else c2+=a[i];
  16.      }
  17.      if(c1 < ) cout << "No " << -c1 << endl;
  18.      else cout << "Yes " << c2 << endl;
  19.      return ;
  20.  }

1093 Count PAT's

统计一下P,A,T。在A位置计算前面有几个P，后面有几个T。

  1.  // 1093 Count PAT's
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  using namespace std;
  5.  
  6.  typedef long long ll;
  7.  const int N =  1e5 + ;
  8.  const ll mod = ;
  9.  char s[N];
  10.  ll sum[N][];
  11.  
  12.  int main() {
  13.      int n;
  14.      ll ans = ;
  15.      scanf("%s", s + );
  16.      n = strlen(s + );
  17.      for(int i = ; i <= n; i++) {
  18.          for(int j = ; j <= ; j++) sum[i][j] = sum[i - ][j];
  19.          if(s[i] == 'P') sum[i][]++;
  20.          else if(s[i] == 'A') sum[i][]++;
  21.          else sum[i][]++;
  22.      }
  23.      for(int i = ; i <= n; i++) {
  24.          if(s[i] == 'A') {
  25.              ans = (ans + (sum[i][] * (sum[n][] - sum[i][])) % mod) % mod;
  26.          }
  27.      }
  28.      printf("%lld\n", ans);
  29.      return ;
  30.  }

1094 The Largest Generation

DFS跑下树，记录下即可。

  1.  // 1094 The Largest Generation
  2.  #include <vector>
  3.  #include <cstdio>
  4.  using namespace std;
  5.  
  6.  const int N = ;
  7.  int n, m, cnt[N];
  8.  vector <int> children[N];
  9.  
  10.  void dfs(int u, int level) {
  11.      for(int i = ; i < children[u].size(); i++) {
  12.          dfs(children[u][i], level + );
  13.      }
  14.      cnt[level]++;
  15.  }
  16.  
  17.  int main() {
  18.      scanf("%d%d", &n, &m);
  19.      while(m--) {
  20.          int r, k, c;
  21.          scanf("%d%d", &r, &k);
  22.          while(k--) {
  23.              scanf("%d", &c);
  24.              children[r].push_back(c);
  25.          }
  26.      }
  27.      dfs(, );
  28.      int mx = , id;
  29.      for(int i = ; i <= n; i++) {
  30.          if(cnt[i] > mx) {
  31.              mx = cnt[i];
  32.              id = i;
  33.          }
  34.      }
  35.      printf("%d %d\n", mx, id);
  36.      return ;
  37.  }

1095 Cars on Campus

按同个车牌号并以时间排序，模拟过去即可。

  1.  // 1095 Cars on Campus
  2.  #include <map>
  3.  #include <vector>
  4.  #include <cstdio>
  5.  #include <cstring>
  6.  #include <iostream>
  7.  #include <algorithm>
  8.  using namespace std;
  9.  
  10.  const int N = 1e4 + ;
  11.  char s[N][], op[];
  12.  int t[N], f[N], id[N];
  13.  int cnt[ *  + ], maxtime;
  14.  map<string, int> m;
  15.  vector <string> ans;
  16.  
  17.  bool cmp(int x, int y) {
  18.      if(!strcmp(s[x], s[y])) {
  19.          return t[x] < t[y];
  20.      } else {
  21.          return strcmp(s[x], s[y]) < ;
  22.      }
  23.  }
  24.  
  25.  int main() {
  26.      int n, k, hh, mm, ss;
  27.      scanf("%d%d", &n, &k);
  28.      for(int i = ; i < n; i++) {
  29.          id[i] = i;
  30.          scanf("%s %d:%d:%d %s", s[i], &hh, &mm, &ss, op);
  31.          t[i] =  * hh +  * mm + ss;
  32.          if(op[] == 'i') f[i] = ;
  33.          else f[i] = ;
  34.      }
  35.      sort(id, id + n, cmp);
  36.      for(int i = ; i < n; i++) {
  37.          if(!strcmp(s[id[i]], s[id[i - ]])) {
  38.              if(f[id[i - ]] ==  && f[id[i]] == ) {
  39.                  cnt[t[id[i - ]]]++;
  40.                  cnt[t[id[i]]]--;
  41.                  m[s[id[i]]] = m[s[id[i]]] + t[id[i]] - t[id[i - ]];
  42.                  maxtime = max(maxtime, m[s[id[i]]]);
  43.              }
  44.          }
  45.      }
  46.      for(int i = ; i <=  * ; i++) cnt[i] += cnt[i - ];
  47.      while(k--) {
  48.          scanf("%d:%d:%d", &hh, &mm, &ss);
  49.          hh =  * hh +  * mm + ss;
  50.          printf("%d\n", cnt[hh]);
  51.      }
  52.      for(int i = ; i < n; i++) {
  53.          if(m[s[i]] == maxtime) {
  54.              ans.push_back(s[i]);
  55.              m[s[i]] = ;
  56.          }
  57.      }
  58.      sort(ans.begin(), ans.end());
  59.      for(int i = ; i < ans.size(); i++) cout << ans[i] << " ";
  60.      hh = maxtime / ; maxtime -=  * hh;
  61.      mm = maxtime / ; maxtime -=  * mm;
  62.      ss = maxtime;
  63.      printf("%02d:%02d:%02d\n", hh, mm, ss);
  64.      return ;
  65.  }

1096 Consecutive Factors

暴力找起点，然后找可以被整除的数。

  1.  // 1096 Consecutive Factors
  2.  #include <iostream>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  typedef long long ll;
  7.  
  8.  int main() {
  9.      ll n, f = , maxlen = ;
  10.      cin >> n;
  11.      for(ll i = ; i * i <= n; i++){
  12.          ll m = , j = i;
  13.          for(; j * j <= n; j++) {
  14.              m *= j;
  15.              if(n % m != ) break;
  16.          }
  17.          if(j - i > maxlen) {
  18.              maxlen = j - i;
  19.              f = i;
  20.          }
  21.      }
  22.      if(f == ) {
  23.          cout <<  << endl << n << endl;
  24.          return ;
  25.      }
  26.      cout << maxlen << endl;
  27.      ll tmp = ;
  28.      cout << f;
  29.      for(ll i = f + ; tmp < n && (i - f) < maxlen; i++) {
  30.          tmp *= i;
  31.          cout << "*" << i;
  32.      }
  33.      return ;
  34.  }

1097 Deduplication on a Linked List

从根跑下树，把仍然保留的移除的都存起来。最后输出即可。

  1.  #include <map>
  2.  #include <vector>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e5 + ;
  9.  int nxt[N], val[N];
  10.  map<int, int> m;
  11.  
  12.  struct node{
  13.      int u, w, v;
  14.  };
  15.  
  16.  vector <node> Remain, Remove;
  17.  
  18.  int main() {
  19.      int root, cur, n;
  20.      scanf("%d %d", &root, &n);
  21.      for(int i = ; i <= n; i++) {
  22.          int u, v, w;
  23.          scanf("%d %d %d", &u, &w, &v);
  24.          nxt[u] = v;
  25.          val[u] = w;
  26.      }
  27.      cur = root;
  28.      while(cur != -) {
  29.          if(!m[abs(val[cur])]) {
  30.              Remain.push_back({cur, val[cur], nxt[cur]});
  31.              m[abs(val[cur])] = ;
  32.          } else {
  33.              Remove.push_back({cur, val[cur], nxt[cur]});
  34.          }
  35.          cur = nxt[cur];
  36.      }
  37.      for(int i = ; i < Remain.size(); i++) {
  38.          if(i != Remain.size() - ) printf("%05d %d %05d\n", Remain[i].u, Remain[i].w, Remain[i + ].u);
  39.          else printf("%05d %d -1\n", Remain[i].u, Remain[i].w);
  40.      }
  41.      for(int i = ; i < Remove.size(); i++) {
  42.          if(i != Remove.size() - ) printf("%05d %d %05d\n", Remove[i].u, Remove[i].w, Remove[i + ].u);
  43.          else printf("%05d %d -1\n", Remove[i].u, Remove[i].w);
  44.      }
  45.      return ;
  46.  }

1098 Insertion or Heap Sort

判断下排序类型，插入排序很容易解决，再扔个堆排序上去即可。

  1.  // 1098 Insertion or Heap Sort
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = ;
  9.  int a[N], b[N];
  10.  
  11.  void solve(int l, int r) {
  12.      int i = , j = ;
  13.      while(j <= r) {
  14.          if(j +  <= r && b[j] < b[j + ]) j++;
  15.          if(b[j] > b[i]) {
  16.              swap(b[j], b[i]);
  17.              i = j;
  18.              j =  * (i + ) -;
  19.          } else {
  20.              break;
  21.          }
  22.      }
  23.  }
  24.  
  25.  int main() {
  26.      int n, k, f = ;
  27.      scanf("%d", &n);
  28.      for(int i = ; i < n; i++) scanf("%d", &a[i]);
  29.      for(int i = ; i < n; i++) scanf("%d", &b[i]);
  30.      for(k = ; k < n -  && b[k] <= b[k + ]; k++) ;
  31.      for(int i = k + ; i < n; i++) {
  32.          if(a[i] != b[i]) {f = ; break;}
  33.      }
  34.      if(f) {
  35.          printf("Insertion Sort\n");
  36.          sort(b, b + k + );
  37.          for(int i = ; i < n; i++){
  38.              printf("%d", b[i]);
  39.              if(i != n - ) printf(" ");
  40.          }
  41.      } else {
  42.          printf("Heap Sort\n");
  43.          int p = n - ;
  44.          while(p >  && b[p - ] <= b[p]) {
  45.              p--;
  46.          }
  47.          swap(b[], b[p]);
  48.          solve(, p - );
  49.          for(int i = ; i < n; i++){
  50.              printf("%d", b[i]);
  51.              if(i != n - ) printf(" ");
  52.          }
  53.      }
  54.      return ;
  55.  }

1099 Build A Binary Search Tree

先排序，中序遍历即为填数的顺序。最后再层次遍历输出。

  1.  // 1099 Build A Binary Search Tree
  2.  #include <queue>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  queue <int> q;
  9.  const int N = ;
  10.  int lch[N], rch[N], val[N], a[N], cnt = ;
  11.  
  12.  void dfs(int u) {
  13.      if(lch[u] != -) dfs(lch[u]);
  14.      val[u] = a[cnt++];
  15.      if(rch[u] != -) dfs(rch[u]);
  16.  }
  17.  
  18.  int main() {
  19.      int n, f = ;
  20.      scanf("%d", &n);
  21.      for(int i = ; i < n; i++) {
  22.          int l, r;
  23.          scanf("%d %d", &l, &r);
  24.          lch[i] = l; rch[i] = r;
  25.      }
  26.      for(int i = ; i < n; i++) scanf("%d", &a[i]);
  27.      sort(a, a + n);
  28.      dfs();
  29.      q.push();
  30.      while(!q.empty()) {
  31.          int cur = q.front();
  32.          q.pop();
  33.          if(!f) f = ;
  34.          else printf(" ");
  35.          printf("%d", val[cur]);
  36.          if(lch[cur] != -) q.push(lch[cur]);
  37.          if(rch[cur] != -) q.push(rch[cur]);
  38.      }
  39.      return ;
  40.  }

1100 Mars Numbers

模拟。

  1.  // 1100 Mars Numbers
  2.  #include <iostream>
  3.  using namespace std;
  4.  
  5.  string s;
  6.  string str[] = {
  7.      "tret", "jan", "feb", "mar", "apr", "may", "jun",
  8.      "jly", "aug", "sep", "oct", "nov", "dec", "tam", "hel",
  9.      "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok",
  10.      "mer", "jou"
  11.  };
  12.  
  13.  void cal1() {
  14.      if(s[] == '') {
  15.          cout << "tret" << endl;
  16.          return ;
  17.      }
  18.      int num = ;
  19.      for(int i = ; i < s.size(); i++) num =  * num + (s[i] - '');
  20.      if(num / ) cout << str[num /  + ];
  21.      if((num / ) && (num % )) cout << " ";
  22.      if(num % ) cout << str[num % ];
  23.      cout << endl;
  24.  }
  25.  
  26.  void cal2() {
  27.      int t1 = , t2 = ;
  28.      string s1 = s.substr(, ), s2 = "";
  29.      if(s.size() > ) s2 = s.substr(, );
  30.      for(int i = ; i <= ; i++) {
  31.          if(str[i] == s1) t1 = i;
  32.          if(str[i] == s2) t2 = i;
  33.      }
  34.      if(t1 > ) cout << (t1 - ) *  + t2 << endl;
  35.      else cout << t1 + t2 << endl;
  36.  }
  37.  
  38.  int main() {
  39.      int n;
  40.      cin >> n;
  41.      getchar();
  42.      for(int i = ; i <= n; i++) {
  43.          getline(cin, s);
  44.          if(s[] >= '' && s[] <= '') cal1();
  45.          else cal2();
  46.      }
  47.      return ;
  48.  }

1101 Quick Sort

用个数组记录下从前往后，对于当前位置的最大值；从后往前，对于当前位置的最小值。最后再遍历一遍数组即可。

  1.  // 1101 Quick Sort
  2.  #include <vector>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  vector <int> ans;
  9.  const int N = 1e5 + ;
  10.  int a[N], pre[N], nxt[N];
  11.  
  12.  int main() {
  13.      int n;
  14.      scanf("%d", &n);
  15.      for(int i = ; i <= n; i++) scanf("%d", &a[i]);
  16.      for(int i = ; i <= n; i++) {
  17.          pre[i] = max(pre[i - ], a[i]);
  18.      }
  19.      nxt[n + ] = 1e9 + ;
  20.      for(int i = n; i >= ; i--) {
  21.          nxt[i] = min(nxt[i + ], a[i]);
  22.      }
  23.      for(int i = ; i <= n; i++) {
  24.          if(pre[i] <= a[i] && nxt[i] >= a[i]) {
  25.              ans.push_back(a[i]);
  26.          }
  27.      }
  28.      printf("%d\n", ans.size());
  29.      sort(ans.begin(), ans.end());
  30.      for(int i = ; i < ans.size(); i++) {
  31.          if(i == ) printf("%d", ans[i]);
  32.          else printf(" %d", ans[i]);
  33.      }
  34.      printf("\n");
  35.      return ;
  36.  }

1102 Invert a Binary Tree

先找下根，中序和层序，把先左后右的顺序调换下。

  1.  // 1102 Invert a Binary Tree
  2.  #include <queue>
  3.  #include <cstdio>
  4.  #include <cstring>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N = ;
  10.  int n, root, f;
  11.  int lch[N], rch[N], fa[N];
  12.  
  13.  void level_order() {
  14.      queue <int> q;
  15.      q.push(root);
  16.      while(!q.empty()) {
  17.          int u = q.front();
  18.          if(!f) f = ;
  19.          else printf(" ");
  20.          printf("%d", u);
  21.          q.pop();
  22.          if(rch[u] != -) q.push(rch[u]);
  23.          if(lch[u] != -) q.push(lch[u]);
  24.      }
  25.      printf("\n");
  26.  }
  27.  
  28.  void in_order(int u) {
  29.      if(rch[u] != -) in_order(rch[u]);
  30.      if(!f) f = ;
  31.      else printf(" ");
  32.      printf("%d", u);
  33.      if(lch[u] != -) in_order(lch[u]);
  34.  }
  35.  
  36.  int main() {
  37.      cin >> n;
  38.      memset(fa, -, sizeof(fa));
  39.      memset(lch, -, sizeof(lch));
  40.      memset(rch, -, sizeof(rch));
  41.      for(int i = ; i < n; i++) {
  42.          char c1, c2;
  43.          cin >> c1 >> c2;
  44.          if(c1 != '-') {
  45.              lch[i] = (c1 - '');
  46.              fa[(c1 - '')] = i;
  47.          }
  48.          if(c2 != '-') {
  49.              rch[i] = (c2 - '');
  50.              fa[(c2 - '')] = i;
  51.          }
  52.      }
  53.      for(int i = ; i < n; i++){
  54.          if(fa[i] == -) root = i;
  55.      }
  56.      f = ;
  57.      level_order();
  58.      f = ;
  59.      in_order(root);
  60.      printf("\n");
  61.      return ;
  62.  }

1103 Integer Factorization

预处理下，再DFS+剪枝。更新下答案。

  1.  // 1103 Integer Factorization
  2.  #include <cstdio>
  3.  #include <iostream>
  4.  #include <algorithm>
  5.  using namespace std;
  6.  
  7.  int a[], b[], ans[];
  8.  int n, m, k, c;
  9.  
  10.  void dfs(int id, int cnt, int val) {
  11.      b[cnt] = id;
  12.      if(cnt > m || val > n) return ;
  13.      if(val == n && cnt == m) {
  14.          bool f = ;
  15.          for(int i = ; i <= m; i++) {
  16.              if(ans[i] != b[i]) {
  17.                  if(b[i] > ans[i]) f = ;
  18.                  break;
  19.              }
  20.          }
  21.          for(int i = ; i <= m; i++) ans[i] = b[i];
  22.          return ;
  23.      }
  24.      for(int i = id; i <= c && val + a[i] <= n; i++) {
  25.          dfs(i, cnt + , val + a[i]);
  26.      }
  27.  }
  28.  
  29.  int POW(int u, int v) {
  30.      int res = ;
  31.      for(int i = ; i <= v; i++) res *= u;
  32.      return res;
  33.  }
  34.  
  35.  int main() {
  36.      scanf("%d%d%d", &n, &m, &k);
  37.      for(int i = ; POW(i, k) <= n; i++) {
  38.          a[i] = POW(i, k);
  39.          c = i;
  40.      }
  41.      for(int i = ; i <= c; i++) dfs(i, , a[i]);
  42.      int res = ;
  43.      for(int i = ; i <= m; i++) res += a[ans[i]];
  44.      if(res == n) {
  45.          printf("%d = %d^%d", n, ans[m], k);
  46.          for(int i = m - ; i >= ; i--) {
  47.              printf(" + %d^%d", ans[i], k);
  48.          }
  49.          printf("\n");
  50.      } else {
  51.          printf("Impossible\n");
  52.      }
  53.      return ;
  54.  }
 
1104 Sum of Number Segments
每个位置会算到的次数为$i * (n - i + 1)$。
  1.  #include <cstdio>
  2.  using namespace std;
  3.  
  4.  const int N = 1e5 + ;
  5.  double a[N];
  6.  
  7.  int main() {
  8.      int n;
  9.      double ans = ;
  10.      scanf("%d", &n);
  11.      for(int i = ; i <= n; i++) scanf("%lf", &a[i]);
  12.      for(int i = ; i <= n; i++) {
  13.          ans += a[i] * i * (n - i + );
  14.      }
  15.      printf("%.2f\n", ans);
  16.      return ;
  17.  }

1105 Spiral Matrix

蛇形填数，注意下细节。

  1.  // 1105 Spiral Matrix
  2.  #include <cmath>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e4 + ;
  9.  int a[N];
  10.  int ans[N][N];
  11.  
  12.  int main() {
  13.      int n, k;
  14.      scanf("%d", &n);
  15.      for(int i = ; i <= n; i++) scanf("%d", &a[i]);
  16.      if(n == ) {
  17.          printf("%d", a[]);
  18.          return ;
  19.      }
  20.      sort(a + , a +  + n);
  21.      int m = sqrt(1.0 * n);
  22.      while(n % m != ) m++;
  23.      if((n / m) > m) m = n/ m;
  24.      //m n/m
  25.      k = n; n = n / m;
  26.      int i = , j = ;
  27.      int U = , D = m, L = , R = n;
  28.      while(k > ) {
  29.          while(k >  && j < R) {
  30.              ans[i][j] = a[k--];
  31.              j++;
  32.          }
  33.          while(k >  && i < D) {
  34.              ans[i][j] = a[k--];
  35.              i++;
  36.          }
  37.          while(k >  && j > U) {
  38.              ans[i][j] = a[k--];
  39.              j--;
  40.          }
  41.          while(k >  && i > L) {
  42.              ans[i][j] = a[k--];
  43.              i--;
  44.          }
  45.          U++; D--; L++; R--;
  46.          i++; j++;
  47.          if(k == ) {
  48.              ans[i][j] = a[k--];
  49.          }
  50.      }
  51.      for(int i = ; i <= m; i++) {
  52.          for(int j = ; j <= n; j++) {
  53.              printf("%d%c", ans[i][j], (j == n) ? '\n' : ' ');
  54.          }
  55.      }
  56.      return ;
  57.  }

1106 Lowest Price in Supply Chain

跑下树。在叶子节点更新下最小值，最后再遍历下叶子节点即可。

  1.  // 1106 Lowest Price in Supply Chain
  2.  #include <cstdio>
  3.  #include <vector>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N = 1e5 + ;
  9.  int n, k, id;
  10.  double p, r, val[N], mi = 1e12;
  11.  vector <int> E[N];
  12.  
  13.  void dfs(int u) {
  14.      if(E[u].size() == ) mi = min(mi, val[u]);
  15.      for(int j = ; j < E[u].size(); j++) {
  16.          int v = E[u][j];
  17.          val[v] = val[u] * (1.0 + r / 100.0);
  18.          dfs(v);
  19.      }
  20.  }
  21.  
  22.  int main() {
  23.      scanf("%d %lf %lf", &n, &p, &r);
  24.      for(int i = ; i < n; i++) {
  25.          scanf("%d", &k);
  26.          for(int j = ; j <= k; j++) {
  27.              scanf("%d", &id);
  28.              E[i].push_back(id);
  29.          }
  30.      }
  31.      val[] = p;
  32.      dfs();
  33.      int cnt = ;
  34.      for(int i = ; i < n; i++) if(val[i] == mi && E[i].size() == ) cnt++;
  35.      printf("%.4f %d\n", mi, cnt);
  36.      return ;
  37.  }

1107 Social Clusters

并查集搞搞。

  1.  // 1107 Social Clusters
  2.  #include <cstdio>
  3.  #include <vector>
  4.  #include <cstring>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N = 1e3 + ;
  10.  vector <int> E[N], ans;
  11.  int fa[N], C[N];
  12.  
  13.  int fi(int x) {
  14.      return x == fa[x] ? x : fa[x] = fi(fa[x]);
  15.  }
  16.  
  17.  void Union(int x, int y) {
  18.      int fx = fi(x), fy = fi(y);
  19.      if(fx != fy) {
  20.          fa[fy] = fx;
  21.      }
  22.  }
  23.  
  24.  int main() {
  25.      bool f = ;
  26.      int n, k, u, cnt = ;
  27.      scanf("%d", &n);
  28.      for(int i = ; i <= n; i++) {
  29.          fa[i] = i;
  30.          scanf("%d:", &k);
  31.          while(k--) {
  32.              scanf("%d", &u);
  33.              E[u].push_back(i);
  34.          }
  35.      }
  36.      for(int i = ; i < N; i++) {
  37.          for(int j = ; j < E[i].size(); j++) {
  38.              u = E[i][j];
  39.              Union(u, E[i][]);
  40.          }
  41.      }
  42.      for(int i = ; i <= n; i++) {
  43.          C[fi(i)]++;
  44.          if(i == fa[i]) cnt++;
  45.      }
  46.      printf("%d\n", cnt);
  47.      for(int i = ; i <= n; i++) {
  48.          if(C[i] != ) ans.push_back(C[i]);
  49.      }
  50.      sort(ans.begin(), ans.end());
  51.      for(int i = ans.size() - ; i >= ; i--) {
  52.          if(i != ans.size() - ) printf(" ");
  53.          printf("%d", ans[i]);
  54.      }
  55.      return ;
  56.  }

1108 Finding Average

注意 一个合法数字 的输出格式。

  1.  // 1108 Finding Average
  2.  #include <cstdio>
  3.  #include <cstring>
  4.  using namespace std;
  5.  
  6.  char str[];
  7.  
  8.  double check() {
  9.      int i = , f = , cnt = ;
  10.      if(str[] == '-') i = , f = -;
  11.      int m = strlen(str);
  12.      for(; i < m; i++) {
  13.          if(str[i] >= '' && str[i] <= '') continue;
  14.          if(str[i] == '.') {
  15.              cnt++;
  16.              if(cnt >= ) return ;
  17.          }
  18.          else return ;
  19.      }
  20.      double res = ;
  21.      i = ;
  22.      if(str[] == '-') i = , f = -;
  23.      for(; i < m; i++) {
  24.          if(str[i] == '.') {
  25.              if(i +  < m) return ;
  26.              if(i +  < m) res += (str[i + ] - '') / 100.0;
  27.              if(i +  < m) res += (str[i + ] - '') / 10.0;
  28.              return f * res;
  29.          }
  30.          res =  * res + 1.0 * (str[i] - '');
  31.      }
  32.      return f * res;
  33.  }
  34.  
  35.  int main() {
  36.      double average = ;
  37.      int n, c = ;
  38.      scanf("%d", &n);
  39.      while(n--) {
  40.          scanf("%s", str);
  41.          double val = check();
  42.          if(val >= - && val <= ) average += check(), c++;
  43.          else{
  44.              printf("ERROR: %s is not a legal number\n", str);
  45.          }
  46.      }
  47.      if(c == ) printf("The average of %d number is %.2f\n", c, average / c);
  48.      else if(c > ) printf("The average of %d numbers is %.2f\n", c, average / c);
  49.      else printf("The average of 0 numbers is Undefined\n");
  50.      return ;
  51.  }

1109 Group Photo

模拟。

  1.  // 1109 Group Photo
  2.  #include <vector>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  struct node {
  9.      int h;
  10.      string name;
  11.  } tmp;
  12.  
  13.  bool cmp(node x, node y) {
  14.      if(x.h == y.h) return x.name < y.name;
  15.      return x.h > y.h;
  16.  }
  17.  
  18.  vector<node> p;
  19.  
  20.  int main() {
  21.      int n, k, per, last, id = ;
  22.      cin >> n >> k;
  23.      per = n / k;
  24.      last = per + (n % k);
  25.      for (int i = ; i <= n; i++) {
  26.          cin >> tmp.name >> tmp.h;
  27.          p.push_back(tmp);
  28.      }
  29.      sort(p.begin(), p.end(), cmp);
  30.      for (int i = ; i < k; i++) {
  31.          int now = per, f = -;
  32.          if(i == ) now = last;
  33.          int pos = now / ;
  34.          vector<string> ans(n);
  35.          while(pos >=  && pos < now) {
  36.              ans[pos] = p[id++].name;
  37.              pos += f;
  38.              if(f < ) {
  39.                  f *= -;
  40.                  f++;
  41.              } else {
  42.                  f *= -;
  43.                  f--;
  44.              }
  45.          }
  46.          for (int j = ; j < now; j++) {
  47.              cout << ans[j];
  48.              if(j != now - ) cout << " ";
  49.              else cout << endl;
  50.          }
  51.          ans.clear();
  52.      }
  53.      return ;
  54.  }

1110 Complete Binary Tree

完全二叉树递归出的最大下标值一定等于总节点数。

  1.  // 1110 Complete Binary Tree
  2.  #include <queue>
  3.  #include <iostream>
  4.  using namespace std;
  5.  
  6.  int fa[], lch[], rch[];
  7.  int n, root, last, maxindex = ;
  8.  
  9.  void dfs(int u, int index) {
  10.      if(u == -) return ;
  11.      if(index > maxindex) {
  12.          maxindex = index;
  13.          last = u;
  14.      }
  15.      dfs(lch[u],  * index);
  16.      dfs(rch[u],  * index + );
  17.  }
  18.  
  19.  int main() {
  20.      cin >> n;
  21.      for(int i = ; i < n; i++) fa[i] = lch[i] = rch[i] = -;
  22.      for(int i = ; i < n; i++) {
  23.          string s1, s2;
  24.          cin >> s1 >> s2;
  25.          if(s1 != "-") {
  26.              fa[atoi(s1.c_str())] = i;
  27.              lch[i] = atoi(s1.c_str());
  28.          }
  29.          if(s2 != "-") {
  30.              fa[atoi(s2.c_str())] = i;
  31.              rch[i] = atoi(s2.c_str());
  32.          }
  33.      }
  34.      for(int i = ; i < n; i++) {
  35.          if(fa[i] == -) root = i;
  36.      }
  37.      dfs(root, );
  38.      if(maxindex == n) {
  39.          cout << "YES " << last << endl;
  40.      } else {
  41.          cout << "NO " << root << endl;
  42.      }
  43.      return ;
  44.  }

1112 Stucked Keyboard

模拟。

  1.  // 1112 Stucked Keyboard
  2.  #include <map>
  3.  #include <iostream>
  4.  using namespace std;
  5.  
  6.  string str;
  7.  map<char, int> ok;
  8.  
  9.  int main() {
  10.      int n, k, c = ;
  11.      cin >> k >> str;
  12.      n = str.size();
  13.      for (int i = ; i < n; i++) {
  14.          if(str[i] == str[i - ]) {
  15.              c++;
  16.          } else {
  17.              if(c % k != ) ok[str[i - ]] = ;
  18.              c = ;
  19.          }
  20.      }
  21.      if(c >  && c % k != ) ok[str[n - ]] = ;
  22.      for (int i = ; i < n; i++) {
  23.          if(ok[str[i]] == ) {
  24.              cout << str[i];
  25.              ok[str[i]] = -;
  26.          }
  27.      }
  28.      cout << endl;
  29.      c = ;
  30.      for (int i = ; i < n; i++) {
  31.          if(str[i] == str[i - ]) {
  32.              c++;
  33.          } else {
  34.              if(c % k == ) {
  35.                  if(ok[str[i - ]] != ) for (int j = ; j <= c / k; j++) cout << str[i - ];
  36.                  else  for (int j = ; j <= c; j++) cout << str[i - ];
  37.              }
  38.              else {
  39.                  for (int j = ; j <= c; j++) cout << str[i - ];
  40.              }
  41.              c = ;
  42.          }
  43.      }
  44.      if(c >  && c % k == ) {
  45.          if(ok[str[n - ]] != ) for (int j = ; j <= c / k; j++) cout << str[n - ];
  46.          else  for (int j = ; j <= c; j++) cout << str[n - ];
  47.      }
  48.      else {
  49.          for (int j = ; j <= c; j++) cout << str[n - ];
  50.      }
  51.      return ;
  52.  }

1113 Integer Set Partition

先排序，再前面加一加，后面加一加。

  1.  // 1113 Integer Set Partition
  2.  #include <iostream>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  typedef long long ll;
  7.  const int N = 1e5 + ;
  8.  ll a[N];
  9.  
  10.  int main() {
  11.      int n, m;
  12.      ll s1 = , s2 = ;
  13.      cin >> n;
  14.      for (int i = ; i <= n; i++) cin >> a[i];
  15.      sort(a + , a +  + n);
  16.      for (int i = n /  + ; i <= n; i++) s1 += a[i];
  17.      for (int i = ; i < n /  + ; i++) s2 += a[i];
  18.      cout << (n % ) << " ";
  19.      cout << abs(s1 - s2) << endl;
  20.      return ;
  21.  }

1114 Family Property

暴力模拟即可。

  1.  // 1114 Family Property
  2.  #include <map>
  3.  #include <vector>
  4.  #include <iostream>
  5.  #include <algorithm>
  6.  using namespace std;
  7.  
  8.  const int N =  + ;
  9.  int fa[N], M[N], A[N];
  10.  int fi(int x) {
  11.      return fa[x] == x ? x : fa[x] = fi(fa[x]);
  12.  }
  13.  
  14.  void Union(int x, int y) {
  15.      if(x == - || y == -) return ;
  16.      int fx = fi(x), fy = fi(y);
  17.      if(fx != fy) {
  18.          fa[fx] = fy;
  19.      }
  20.  }
  21.  
  22.  void init() {
  23.      for(int i = ; i < N; i++) fa[i] = i;
  24.  }
  25.  
  26.  int cnt = ;
  27.  vector<int> V, E[N];
  28.  map<int, int> mp;
  29.  
  30.  struct node {
  31.      int ansID, ansM;
  32.      double avg1, avg2;
  33.  }ans[N];
  34.  
  35.  map<int, int> vis;
  36.  
  37.  bool cmp(node x, node y) {
  38.      if(x.avg2 == y.avg2) return x.ansID < y.ansID;
  39.      return x.avg2 > y.avg2;
  40.  }
  41.  
  42.  int main() {
  43.      init();
  44.      int n;
  45.      cin >> n;
  46.      for (int i = ; i <= n; i++) {
  47.          int id, father, mother, child, k;
  48.          cin >> id >> father >> mother >> k;
  49.          Union(id, father);
  50.          if(father != -) V.push_back(father);
  51.          Union(id, mother);
  52.          if(mother != -) V.push_back(mother);
  53.          while (k--) {
  54.              cin >> child;
  55.              if(child != -) V.push_back(child);
  56.              Union(id, child);
  57.          }
  58.          V.push_back(id);
  59.          cin >> M[id] >> A[id];
  60.      }
  61.      for (int i = ; i < V.size(); i++) {
  62.          int u = V[i];
  63.          if(fa[u] == u && !mp[u]) {
  64.              cnt++;
  65.              mp[u] = cnt;
  66.          }
  67.      }
  68.      for (int i = ; i < V.size(); i++) {
  69.          int u = V[i];
  70.          if(!vis[u]) E[mp[fi(u)]].push_back(u), vis[u] = ;
  71.      }
  72.      for (int i = ; i <= cnt; i++) {
  73.          int minid = ;
  74.          double sum1 = ;
  75.          double sum2 = ;
  76.          for(int j = ; j < E[i].size(); j++) {
  77.              int v = E[i][j];
  78.              minid = min(minid, v);
  79.              sum1 += M[v];
  80.              sum2 += A[v];
  81.          }
  82.          ans[i].ansID = minid;
  83.          ans[i].ansM = E[i].size();
  84.          ans[i].avg1 = sum1 / E[i].size();
  85.          ans[i].avg2 = sum2 / E[i].size();
  86.      }
  87.      sort(ans + , ans +  + n, cmp);
  88.      cout << cnt << endl;
  89.      for (int i = ; i <= cnt; i++) {
  90.          printf("%04d %d %.3f %.3f\n", ans[i].ansID, ans[i].ansM, ans[i].avg1, ans[i].avg2);
  91.      }
  92.      return ;
  93.  }

1115 Counting Nodes in a BST

用链表实现二叉搜索树的插入。构建完树后，DFS一遍，在对应深度更新节点数。

  1.  // 1115 Counting Nodes in a BST
  2.  #include <iostream>
  3.  using namespace std;
  4.  
  5.  struct node {
  6.      int val;
  7.      struct node *left, *right;
  8.  };
  9.  
  10.  const int N =  + ;
  11.  int num[N], maxd = -;
  12.  
  13.  node *build(node *root, int v) {
  14.      if(root == NULL) {
  15.          root = new node();
  16.          root -> val = v;
  17.          root -> left = root -> right = NULL;
  18.      } else if(v > root->val) {
  19.          root->right = build(root->right, v);
  20.      } else {
  21.          root->left = build(root->left, v);
  22.      }
  23.      return root;
  24.  }
  25.  
  26.  void dfs(node *root, int dp) {
  27.      if(root == NULL) {
  28.          maxd = max(maxd, dp);
  29.          return ;
  30.      }
  31.      num[dp]++;
  32.      dfs(root->left, dp + );
  33.      dfs(root->right, dp + );
  34.  }
  35.  
  36.  int main() {
  37.      int n, k;
  38.      node *root = NULL;
  39.      cin >> n;
  40.      while(n--) {
  41.          cin >> k;
  42.          root = build(root, k);
  43.      }
  44.      dfs(root, );
  45.      cout << num[maxd - ] << " + " << num[maxd - ] << " = " << num[maxd - ] + num[maxd - ] << endl;
  46.      return ;
  47.  }

1116 Come on! Let's C

模拟。

  1.  // 1116 Come on! Let's C
  2.  #include <map>
  3.  #include <iostream>
  4.  using namespace std;
  5.  
  6.  map<int, int> Rank;
  7.  
  8.  bool check(int x) {
  9.      if(x < ) return false;
  10.      for(int i = ; i * i <= x; i++) {
  11.          if(x % i == ) return false;
  12.      }
  13.      return true;
  14.  }
  15.  
  16.  int main() {
  17.      int n, k;
  18.      cin >> n;
  19.      for (int i = ; i <= n; i++) {
  20.          cin >> k;
  21.          Rank[k] = i;
  22.      }
  23.      cin >> n;
  24.      while (n--) {
  25.          cin >> k;
  26.          printf("%04d: ", k);
  27.          if(Rank[k] == -) cout << "Checked" << endl;
  28.          else if(Rank[k] == ) cout << "Are you kidding?" << endl;
  29.          else if(Rank[k] == ) cout << "Mystery Award" << endl, Rank[k] = -;
  30.          else if(check(Rank[k])) cout << "Minion" << endl, Rank[k] = -;
  31.          else cout << "Chocolate" << endl, Rank[k] = -;
  32.      }
  33.      return ;
  34.  }

1117 Eddington Number

前缀和

  1.  // 1117 Eddington Number
  2.  #include <iostream>
  3.  #include <algorithm>
  4.  using namespace std;
  5.  
  6.  const int N = 1e6 + ;
  7.  int d[N];
  8.  
  9.  int main() {
  10.      int n, k, ans = ;
  11.      cin >> n;
  12.      for(int i = ; i <= n; i++) {
  13.          cin >> k;
  14.          d[k]++;
  15.      }
  16.      for(int i = ; i < N; i++) d[i] += d[i - ];
  17.      for(int i = ; i < N; i++) {
  18.          if((n - d[i]) >= i) ans = max(ans, i);
  19.      }
  20.      cout << ans << endl;
  21.      return ;
  22.  }

1118 Birds in Forest

并查集

  1.  // 1118 Birds in Forest
  2.  #include <cstdio>
  3.  #include <vector>
  4.  using namespace std;
  5.  
  6.  const int N = 1e4 + ;
  7.  bool vis[N];
  8.  int fa[N], cnt = ;
  9.  vector <int> birds;
  10.  
  11.  int fi(int x) {
  12.      return x == fa[x] ? x : fa[x] = fi(fa[x]);
  13.  }
  14.  
  15.  void Union(int x, int y) {
  16.      int fx = fi(x), fy = fi(y);
  17.      if(fx != fy) {
  18.          fa[fx] = fy;
  19.      }
  20.  }
  21.  
  22.  void init() {
  23.      for (int i = ; i < N; i++) fa[i] = i;
  24.  }
  25.  
  26.  int main() {
  27.      init();
  28.      int m, n, k, u, v;
  29.      scanf("%d", &n);
  30.      while (n--) {
  31.          scanf("%d", &k);
  32.          for (int i = ; i <= k; i++) {
  33.              scanf("%d", &v);
  34.              if(!vis[v]) vis[v] = , birds.push_back(v);
  35.              if(i == ) u = v;
  36.              else Union(u, v);
  37.          }
  38.      }
  39.      for (int i = ; i < birds.size(); i++) {
  40.          u = birds[i];
  41.          if(fa[u] == u) cnt++;
  42.      }
  43.      printf("%d %d\n", cnt, birds.size());
  44.      scanf("%d", &m);
  45.      while (m--) {
  46.          scanf("%d %d", &u, &v);
  47.          if(fi(u) == fi(v)) printf("Yes\n");
  48.          else printf("No\n");
  49.      }
  50.      return ;
  51.  }

1119 Pre- and Post-order Traversals 

通过前序和后序先建棵树。（题目中说明可以随意建棵符合的树）再中序遍历即可。

  1.  #include <bits/stdc++.h>
  2.  using namespace std;
  3.  
  4.  const int maxn=;
  5.  int preOrder[maxn], postOrder[maxn];
  6.  bool isUnique=true;
  7.  
  8.  struct Node{
  9.      int left=-,right=-;
  10.  }node[maxn];
  11.  
  12.  void build(int preL,int preR,int postL,int postR){
  13.      if(preL>=preR){
  14.          return;
  15.      }
  16.      int fa=preL;
  17.      int val=preOrder[preL+];
  18.      int postIdx;
  19.      for(int i=postL;i<postR;i++){
  20.          if(val==postOrder[i]){
  21.              postIdx=i;
  22.              break;
  23.          }
  24.      }
  25.      int num=postIdx-postL;
  26.      if(preR-preL-==num){
  27.          isUnique=false;
  28.      }
  29.      node[fa].left=preL+;
  30.      build(preL+,preL+num+,postL,postIdx);
  31.      if(preR-preL->num){
  32.          node[fa].right=preL+num+;
  33.          build(preL+num+,preR,postIdx+,postR-);
  34.      }
  35.  }
  36.  
  37.  bool first=true;
  38.  
  39.  void inOrder(int root){
  40.      if(root==-){
  41.          return;
  42.      }
  43.      inOrder(node[root].left);
  44.      if(first){
  45.          first=false;
  46.          printf("%d",preOrder[root]);
  47.      }
  48.      else
  49.          printf(" %d",preOrder[root]);
  50.      inOrder(node[root].right);
  51.  }
  52.  
  53.  int main(){
  54.      int n;
  55.      scanf("%d",&n);
  56.      for(int i=;i<=n;i++) scanf("%d",&preOrder[i]);
  57.      for(int i=;i<=n;i++) scanf("%d",&postOrder[i]);
  58.      build(,n,,n);
  59.      if(isUnique) printf("Yes\n");
  60.      else printf("No\n");
  61.      inOrder();
  62.      printf("\n");
  63.      return ;
  64.  }

1120 Friend Numbers

模拟

  1.  // 1120 Friend Numbers
  2.  #include <set>
  3.  #include <cstdio>
  4.  #include <iostream>
  5.  using namespace std;
  6.  
  7.  const int N = 1e4 + ;
  8.  int a[N], b[N];
  9.  
  10.  int cal(int x) {
  11.      int sum1 = ;
  12.      while (x) {
  13.          sum1 += x % ;
  14.          x /= ;
  15.      }
  16.      return sum1;
  17.  }
  18.  
  19.  int check(int x, int y) {
  20.      if (x == y) return x;
  21.      return ;
  22.  }
  23.  
  24.  set<int> ans;
  25.  
  26.  int main() {
  27.      int n;
  28.      bool f = ;
  29.      scanf("%d", &n);
  30.      for (int i = ; i <= n; i++) scanf("%d", &a[i]), b[i] = cal(a[i]);
  31.      for (int i = ; i <= n; i++) {
  32.          for (int j = i; j <= n; j++) {
  33.              int val = check(b[i], b[j]);
  34.              if (val) ans.insert(val);
  35.          }
  36.      }
  37.      printf("%d\n", ans.size());
  38.      for (auto x : ans) {
  39.          if (!f) f = ;
  40.          else printf(" ");
  41.          printf("%d", x);
  42.      }
  43.      return ;
  44.  }

1121 Damn Single

模拟

  1.  // 1121 Damn Single
  2.  #include <map>
  3.  #include <vector>
  4.  #include <cstdio>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  map<int, int> m, vis;
  10.  vector<int> p;
  11.  const int N = 1e4 + ;
  12.  int a[N];
  13.  
  14.  int main() {
  15.      int n, x, y;
  16.      scanf("%d", &n);
  17.      while (n--) {
  18.          scanf("%d %d", &x, &y);
  19.          x++; y++;
  20.          m[x] = y;
  21.          m[y] = x;
  22.      }
  23.      scanf("%d", &n);
  24.      for (int i = ; i <= n; i++) {
  25.          scanf("%d", &a[i]);
  26.          a[i]++;
  27.          vis[a[i]] = ;
  28.      }
  29.      for (int i = ; i <= n; i++) {
  30.          if (vis[ m[a[i]] ]) continue;
  31.          p.push_back(a[i]);
  32.      }
  33.      sort(p.begin(), p.end());
  34.      printf("%d\n", p.size());
  35.      for (int i = ; i < p.size(); i++) {
  36.          if (i != ) printf(" ");
  37.          printf("%05d", p[i] - );
  38.      }
  39.      return ;
  40.  }

1122 Hamiltonian Cycle

模拟

  1.  // 1122 Hamiltonian Cycle
  2.  #include <set>
  3.  #include <cstdio>
  4.  #include <cstring>
  5.  #include <iostream>
  6.  #include <algorithm>
  7.  using namespace std;
  8.  
  9.  const int N = ;
  10.  int MAP[N][N];
  11.  set<int> s;
  12.  
  13.  int main() {
  14.      memset(MAP, -, sizeof(MAP));
  15.      int n, m, k;
  16.      scanf("%d %d", &n, &m);
  17.      for (int i = ; i <= m; i++) {
  18.          int u, v;
  19.          scanf("%d %d", &u, &v);
  20.          MAP[u][v] = ; MAP[v][u] = ;
  21.      }
  22.      scanf("%d", &k);
  23.      while (k--) {
  24.          bool f = ;
  25.          int u, v, root;
  26.          scanf("%d", &m);
  27.          scanf("%d", &root);
  28.          s.insert(root);
  29.          u = root;
  30.          for (int i = ; i < m; i++) {
  31.              scanf("%d", &v);
  32.              s.insert(v);
  33.              if (MAP[u][v] == -) f = ;
  34.              u = v;
  35.          }
  36.          if (root != u || s.size() != n || m != n + ) f = ;
  37.          if (!f) printf("NO\n");
  38.          else printf("YES\n");
  39.          s.clear();
  40.      }
  41.      return ;
  42.  }

PAT 甲级真题题解（63-120）的更多相关文章

  1. PAT 甲级真题题解（1-62）

    准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #in ...

  2. PAT 甲级真题题解（121-155）

    1121 Damn Single 模拟 // 1121 Damn Single #include <map> #include <vector> #include <cs ...

  3. 1080 Graduate Admission——PAT甲级真题

    1080 Graduate Admission--PAT甲级练习题 It is said that in 2013, there were about 100 graduate schools rea ...

  4. PAT甲题题解-1107. Social Clusters (30)-PAT甲级真题（并查集）

    题意:有n个人,每个人有k个爱好,如果两个人有某个爱好相同,他们就处于同一个集合.问总共有多少个集合,以及每个集合有多少人,并按从大到小输出. 很明显,采用并查集.vis[k]标记爱好k第一次出现的人 ...

  5. PAT甲题题解-1111. Online Map (30)-PAT甲级真题（模板题，两次Dijkstra，同时记下最短路径）

    题意:给了图,以及s和t,让你求s到t花费的最短路程.最短时间,以及输出对应的路径.   对于最短路程,如果路程一样,输出时间最少的. 对于最短时间,如果时间一样,输出节点数最少的.   如果最短路程 ...

  6. PAT甲级真题及训练集

    正好这个"水水"的C4来了 先把甲级刷完吧.(开玩笑-2017.3.26) 这是一套"伪题解". wacao 刚才登出账号测试一下代码链接,原来是看不到..有空 ...

  7. 1018 Public Bike Management (30分) PAT甲级真题 dijkstra + dfs

    前言: 本题是我在浏览了柳神的代码后,记下的一次半转载式笔记,不经感叹柳神的强大orz,这里给出柳神的题解地址:https://blog.csdn.net/liuchuo/article/detail ...

  8. PAT甲级真题 A1025 PAT Ranking

    题目概述:Programming Ability Test (PAT) is organized by the College of Computer Science and Technology o ...

  9. PAT 甲级真题

    1019. General Palindromic Number 题意:求数N在b进制下其序列是否为回文串,并输出其在b进制下的表示. 思路:模拟N在2进制下的表示求法,“除b倒取余”,之后判断是否回 ...

随机推荐

  1. AngularJS 截取字符串

    参考文章:https://blog.csdn.net/u010234516/article/details/54631525 //过滤器 app.filter('textLengthSet', fun ...

  2. sql 修改、更新、替换 某个字段的部分内容（转载）

    来源:https://blog.csdn.net/jiangnanqbey/article/details/81304834 1*.需求 将表(Ws_FormMain)的字段(order_Number ...

  3. Hibernate框架_搭建第一个Hibernate框架

    一.eclipse搭建 A.创建动态web项目 New-->Dynamic web project(web project) B.导入jar包 1.数据库驱动包 2.hibernate开发必须j ...

  4. phpstorm本地怎么上传到服务器

    连接服务器 菜单栏找到[工具/Tools]->[Deployment/部署]->[Confinguration…/配置…]. 点加号(+),添加一台服务器,填写名称,选择类型为SFTP,点 ...

  5. 用Python写一个贪吃蛇

    最近在学Python,想做点什么来练练手,命令行的贪吃蛇一般是C的练手项目,但是一时之间找不到别的,就先做个贪吃蛇来练练简单的语法. 由于Python监听键盘很麻烦,没有C语言的kbhit(),所以这 ...

  6. typescript中类的继承

    typescript中类的继承用到的是:extends和super 先看一下typescript中类的写法: class Demo{ //类的属性 name:string; age:number; / ...

  7. mssql sqlserver updatetext关键字应用简介说明

    摘要: 下文讲述updatetext的功能及举例说明 实验环境:sql server 2008 R2 updatetext关键字功能及语法说明 updatetext功能说明: updatetext的功 ...

  8. MySQL基础知识-安装MySQL

    前导: 昨天去参加了一个面试,公司不太大,是一家日资企业,在国内有几家分公司,面试官问到了MySQL的基本操作和性能优化,说了一大堆,倒是比较轻松的过了,但是面试结束之后,想了一下,基本操作忘的还是挺 ...

  9. Windows服务器如何查看共享目录信息

    查看Windows服务器上的共享目录的相关信息,可以使用两种方式: 1:命令net share 查看: 2:通过计算机管理的Shared Folders查看

  10. jquery中Json操作

    在开发中,我们有可能拿到的不是全的json,而是一部分json格式的数据,这个时候我们需要将其强转为json对象 第一种方法:使用jquery中的$.parseJSON(),但是它对json数据格式的 ...