$$2015-2016\ ACM-ICPC,\ NEERC,\ Northern\ Subregional\ Contest$$

\(A.Alex\ Origami\ Squares\)

签到

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. int main(){
  5.     int a,b;
  6.     freopen("alex.in","r",stdin);
  7.     cin >> a >> b;
  8.     freopen("alex.out","w",stdout);
  9.     double res = max(max(min(a/3.0,b/1.0),min(a/1.0,b/3.0)),min(a/2.0,b/2.0));
  10.     cout << res << endl;
  11.     return 0;
  12. }

\(B.Black\ and\ White\)

构造，把多的框在少的里面即可

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 1e5+7;
  6. int n,m,len;
  7. char tile[MAXN][5];
  8. int main(){
  9.     #ifdef ONLINE_JUDGE
  10.     freopen("black.in","r",stdin);
  11.     #endif
  12.     ____();
  13.     cin >> n >> m;
  14.     #ifdef ONLINE_JUDGE
  15.     freopen("black.out","w",stdout);
  16.     #endif
  17.     if(m==0){
  18.         cout << 1 << ' ' << 1 << endl << '@' << endl;
  19.         return 0;
  20.     }
  21.     if(n==0){
  22.         cout << 1 << ' ' << 1 << endl << '.' << endl;
  23.         return 0;
  24.     }
  25.     char x='@',y='.';
  26.     if(n<m){
  27.         swap(n,m);
  28.         swap(x,y);
  29.     }
  30.     len = 1;
  31.     if(n>m){
  32.         tile[len][1] = tile[len][2] = tile[len][3] = y;
  33.         len++, m--;
  34.     }
  35.     while(n>m){
  36.         tile[len][2] = x;
  37.         tile[len][1] = tile[len][3] = y;
  38.         len++, n--;
  39.         tile[len][1] = tile[len][2] = tile[len][3] = y;
  40.         len++;
  41.     }
  42.     while(n){
  43.         tile[len][1] = tile[len][2] = tile[len][3] = x;
  44.         len++, n--;
  45.         tile[len][1] = tile[len][2] = tile[len][3] = y;
  46.         len++;
  47.     }
  48.     if(tile[len][1]=='\0') len--;
  49.     cout << len << ' ' << 3 << endl;
  50.     for(int i = 1; i <= len; i++) cout << (tile[i]+1) << endl;
  51.     return 0;
  52. }

\(C.Concatenation\)

答案就是总数量-右串从\(1\)到\(lenr-1\)各个字符在左串\(2~lenl\)之间出现的次数

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 3e5+7;
  6. char s[MAXN],t[MAXN];
  7. int len1,len2;
  8. int main(){
  9.     #ifdef ONLINE_JUDGE
  10.     freopen("concatenation.in","r",stdin);
  11.     #endif
  12.     scanf("%s %s",s+1,t+1);
  13.     #ifdef ONLINE_JUDGE
  14.     freopen("concatenation.out","w",stdout);
  15.     #endif
  16.     len1 = strlen(s+1);
  17.     len2 = strlen(t+1);
  18.     int_fast64_t res = len1*(int_fast64_t)len2;
  19.     map<char,int> mp;
  20.     for(int i = 2; i <= len1; i++) mp[s[i]]++;
  21.     for(int i = 1; i < len2; i++) res -= mp[t[i]];
  22.     cout << res << endl;
  23.     return 0;
  24. }

\(D.Distribution\ in\ Metagonia\)

考虑构造时先把质因子\(2\)，\(3\)提出来，然后得到一个奇数，把奇数拆分为\(3^k·p\)，继续对\(p\)拆分直到\(p\)的因子只含\(2\)和\(3\)为止

递归构造即可

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. using LL = int_fast64_t;
  6. int t;
  7. LL n[1111];
  8. void gao(LL m, LL mul, vector<LL> &vec){
  9.     if(m==1){
  10.         vec.emplace_back(mul);
  11.         return;
  12.     }
  13.     if(m%3!=0&&m%2!=0){
  14.         LL tp = 1;
  15.         while(tp*3<=m) tp *= 3;
  16.         gao(tp,mul,vec);
  17.         gao(m-tp,mul,vec);
  18.         return;
  19.     }
  20.     LL sum = 1;
  21.     while(m%2==0){
  22.         sum *= 2;
  23.         m >>= 1;
  24.     }
  25.     while(m%3==0){
  26.         sum *= 3;
  27.         m /= 3;
  28.     }
  29.     gao(m,sum*mul,vec);
  30. }
  31. void solve(LL m){
  32.     vector<LL> vec;
  33.     gao(m,1,vec);
  34.     cout << vec.size() << endl;
  35.     for(LL x : vec) cout << x << ' '; cout << endl;
  36. }
  37. int main(){
  38.     #ifdef ONLINE_JUDGE
  39.     freopen("distribution.in","r",stdin);
  40.     freopen("distribution.out","w",stdout);
  41.     #endif
  42.     ____();
  43.     cin >> t;
  44.     for(int i = 1; i <= t; i++) cin >> n[i];
  45.     for(int i = 1; i <= t; i++) solve(n[i]);
  46.     return 0;
  47. }

\(E.Easy\ Arithmetic\)

打记号标记前导零、上一位是否是符号、当前正负值然后判断即可

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 3333;
  6. char s[MAXN],tar[MAXN];
  7. int main(){
  8.     #ifdef ONLINE_JUDGE
  9.     freopen("easy.in","r",stdin);
  10.     #endif
  11.     ____();
  12.     cin >> (s+1);
  13.     int len = strlen(s+1);
  14.     #ifdef ONLINE_JUDGE
  15.     freopen("easy.out","w",stdout);
  16.     #endif
  17.     int c = 1, ptr = 1;
  18.     bool tag = true, lead = true, op = true;
  19.     while(ptr<=len){
  20.         if(!isdigit(s[ptr])){
  21.             if(s[ptr]=='-') tag = false;
  22.             else if(s[ptr]=='+') tag = true;
  23.             tar[c++] = s[ptr];
  24.             op = true;
  25.         }
  26.         else{
  27.             if(op){
  28.                 op = false;
  29.                 tar[c++] = s[ptr];
  30.                 lead = (s[ptr]=='0');
  31.             }
  32.             else{
  33.                 if(!tag){
  34.                     tar[c++] = '+';
  35.                     tag = true;
  36.                     tar[c++] = s[ptr];
  37.                     lead = (s[ptr]=='0');
  38.                 }
  39.                 else{
  40.                     if(lead){
  41.                         tar[c++] = '+';
  42.                         lead = (s[ptr]=='0');
  43.                         tar[c++] = s[ptr];
  44.                     }
  45.                     else tar[c++] = s[ptr];
  46.                 }
  47.             }
  48.         }
  49.         ptr++;
  50.     }
  51.     cout << (tar+1) << endl;
  52.     return 0;
  53. }

\(F.Fygon\)

\(G.Graph\)

友情链接

大顶堆小顶堆维护，做法有点妙

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 1e5+7;
  6. int n,m,k,pre,ord[MAXN],cur,deg[MAXN];
  7. vector<int> G[MAXN];
  8. set<int> S;
  9. set<int,greater<int>> done;
  10. vector<pair<int,int>> edge;
  11. void toposort(int u){
  12.     ord[++cur] = u;
  13.     pre = u;
  14.     for(int v : G[u]) if(--deg[v]==0) S.insert(v);
  15. }
  16. int main(){
  17.     #ifdef ONLINE_JUDGE
  18.     freopen("graph.in","r",stdin);
  19.     freopen("graph.out","w",stdout);
  20.     #endif
  21.     scanf("%d %d %d",&n,&m,&k);
  22.     for(int i = 1; i <= m; i++){
  23.         int u, v;
  24.         scanf("%d %d",&u,&v);
  25.         G[u].emplace_back(v);
  26.         deg[v]++;
  27.     }
  28.     for(int i = 1; i <= n; i++) if(deg[i]==0) S.insert(i);
  29.     while(cur<n){
  30.         if(S.empty()){          //自己安排下一个节点顺序
  31.             int u = *done.begin();
  32.             done.erase(u);
  33.             edge.emplace_back(make_pair(pre,u));
  34.             toposort(u);
  35.         }
  36.         else if(!k||(S.size()==1&&(done.empty()||*S.begin()>*done.begin()))){
  37.             int u = *S.begin();
  38.             S.erase(u);
  39.             toposort(u);
  40.         }
  41.         else{
  42.             k--;
  43.             done.insert(*S.begin());
  44.             S.erase(S.begin());
  45.         }
  46.     }
  47.     for(int i = 1; i <= n; i++) printf("%d ",ord[i]); puts("");
  48.     printf("%d\n",edge.size());
  49.     for(auto p : edge) printf("%d %d\n",p.first,p.second);
  50.     return 0;
  51. }

\(H.Hash\ Code\ Hacker\)

每次从后一位向前进一位，后一位\(y\)变成\(Z\)，前一位的\(x\)变成\(y\)，重复操作即可

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 1111;
  6. char s[MAXN];
  7. int k;
  8. int main(){
  9.     #ifdef ONLINE_JUDGE
  10.     freopen("hash.in","r",stdin);
  11.     #endif
  12.     cin >> k;
  13.     #ifdef ONLINE_JUDGE
  14.     freopen("hash.out","w",stdout);
  15.     #endif
  16.     for(int i = 1; i <= 999; i++) s[i] = 'x';
  17.     s[1000] = 'y';
  18.     cout << (s+1) << endl;
  19.     for(int i = 1; i < k; i++){
  20.         s[1000+1-i] = 'Z';
  21.         s[1000-i] = 'y';
  22.         cout << (s+1) << endl;
  23.     }
  24.     return 0;
  25. }

\(I.Insider's\ Information\)

\(J.Journey\ to\ the\ "The World's Start"\)

二分答案之后线段树维护DP求最小耗时

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 1e5+7;
  6. using LL = int_fast64_t;
  7. const LL INF = 0x3f3f3f3f3f3f3f3f;
  8. int n,t,p[MAXN],cost[MAXN];
  9. struct SegmentTree{
  10.     LL minn[MAXN<<2];
  11.     int l[MAXN<<2],r[MAXN<<2];
  12.     #define ls(rt) ((rt) << 1)
  13.     #define rs(rt) ((rt) << 1 | 1)
  14.     void pushup(int rt){ minn[rt] = min(minn[ls(rt)],minn[rs(rt)]); }
  15.     void build(int L, int R, int rt){
  16.         l[rt] = L, r[rt] = R;
  17.         if(L+1==R) return;
  18.         int mid = (L+R) >> 1;
  19.         build(L,mid,ls(rt)); build(mid,R,rs(rt));
  20.         pushup(rt);
  21.     }
  22.     void update(int pos, int rt, LL x){
  23.         if(l[rt]+1==r[rt]) {
  24.             minn[rt] = x;
  25.             return;
  26.         }
  27.         int mid = (l[rt]+r[rt]) >> 1;
  28.         if(pos<mid) update(pos,ls(rt),x);
  29.         else update(pos,rs(rt),x);
  30.         pushup(rt);
  31.     }
  32.     LL qmin(int L, int R, int rt){
  33.         if(l[rt]>=R||L>=r[rt]) return INF;
  34.         if(L<=l[rt] && r[rt]<=R) return minn[rt];
  35.         return min(qmin(L,R,ls(rt)),qmin(L,R,rs(rt)));
  36.     }
  37. }ST;
  38. bool check(int mid){
  39.     ST.update(1,1,-1);
  40.     for(int i = 2; i <= n; i++) ST.update(i,1,ST.qmin(max(1,i-mid),i,1)+cost[i]);
  41.     return ST.qmin(n,n+1,1)+n<=t;
  42. }
  43. int solve(){
  44.     int l = 1, r = n-1;
  45.     ST.build(1,n+1,1);
  46.     while(l<=r){
  47.         int mid = (l+r)>>1;
  48.         if(check(mid)) r = mid - 1;
  49.         else l = mid + 1;
  50.     }
  51.     return p[l];
  52. }
  53. int main(){
  54.     #ifdef ONLINE_JUDGE
  55.     freopen("journey.in","r",stdin);
  56.     freopen("journey.out","w",stdout);
  57.     #endif
  58.     scanf("%d %d",&n,&t);
  59.     for(int i = 1; i < n; i++) scanf("%d",&p[i]);
  60.     for(int i = n-2; i >= 1; i--) p[i] = min(p[i],p[i+1]);
  61.     for(int i = 2; i < n; i++) scanf("%d",&cost[i]);
  62.     printf("%d\n",solve());
  63.     return 0;
  64. }

\(K.Kingdom\ Trip\)

\(L.Lucky\ Chances\)

签到，暴力

  1. //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
  5. const int MAXN = 111;
  6. const int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
  7. int n,m,A[MAXN][MAXN];
  8. int main(){
  9.     #ifdef ONLINE_JUDGE
  10.     freopen("lucky.in","r",stdin);
  11.     #endif
  12.     ____();
  13.     cin >> n >> m;
  14.     for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> A[i][j];
  15.     #ifdef ONLINE_JUDGE
  16.     freopen("lucky.out","w",stdout);
  17.     #endif
  18.     int res = 0;
  19.     for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++){
  20.         for(int d = 0; d < 4; d++){
  21.             bool ok = true;
  22.             for(int x = i+dir[d][0], y = j + dir[d][1]; x>0&&x<=n&&y>0&&y<=m; x+=dir[d][0],y+=dir[d][1]) if(A[x][y]>=A[i][j]) ok = false;
  23.             res+=ok;
  24.         }
  25.     }
  26.     cout << res << endl;
  27.     return 0;
  28. }

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