最近正在学AC自动机，按照惯例需要刷一套kuangbin的AC自动机专题巩固

在网上看过很多模板，感觉kuangbin大神的模板最为简洁，于是就选择了用kuangbin大神的模板。

AC自动机其实就是字典树和KMP的结合，然后去思考一下KMP的原理，然后就是在字典树上实现KMP

这里最重要的思想可能就是fail的思想，就像KMP一样，匹配失败后，有一个next的数组去回溯(最长公共前缀后缀)

如何理解了KMP的话，感觉这个不会很难理解，字典树是一个非常简单的东西就不用讲了吧。

HDU - 2222,HDU - 2896,HDU - 3065,ZOJ - 3430

这是我总结出来的AC自动机解决的第一类问题，求文本串和模式串的关系(模式串出现几次之类的问题)

我把这几题放一起写了算了，因为套路一样随便改一下就好了

Keywords Search HDU - 2222

询问有多少个模式串出现在了文本串里面。

将模式串插入Trie树中，然后直接查询就好了。

  1.  #include <set>
  2.  #include <map>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <cstdio>
  7.  #include <string>
  8.  #include <vector>
  9.  #include <time.h>
  10.  #include <cstring>
  11.  #include <iostream>
  12.  #include <algorithm>
  13.  #include <unordered_map>
  14.  
  15.  #define  pi acos(-1.0)
  16.  #define  eps 1e-9
  17.  #define  fi first
  18.  #define  se second
  19.  #define  rtl   rt<<1
  20.  #define  rtr   rt<<1|1
  21.  #define  bug               printf("******\n")
  22.  #define  mem(a, b)         memset(a,b,sizeof(a))
  23.  #define  name2str(x)       #x
  24.  #define  fuck(x)           cout<<#x" = "<<x<<endl
  25.  #define  sf(n)             scanf("%d", &n)
  26.  #define  sff(a, b)         scanf("%d %d", &a, &b)
  27.  #define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
  28.  #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
  29.  #define  pf                printf
  30.  #define  FIN               freopen("data.txt","r",stdin)
  31.  #define  gcd(a, b)         __gcd(a,b)
  32.  #define  lowbit(x)         x&-x
  33.  #define  IO                iOS::sync_with_stdio(false)
  34.  
  35.  using namespace std;
  36.  typedef long long LL;
  37.  typedef unsigned long long ULL;
  38.  const int maxn = 1e6 + ;
  39.  const int maxm = 8e6 + ;
  40.  const int INF = 0x3f3f3f3f;
  41.  const int mod = 1e9 + ;
  42.  
  43.  struct Aho_Corasick {
  44.      int next[][], fail[], End[];
  45.      int root, cnt;
  46.  
  47.      int newnode() {
  48.          for (int i = ; i < ; i++) next[cnt][i] = -;
  49.          End[cnt++] = ;
  50.          return cnt - ;
  51.      }
  52.  
  53.      void init() {
  54.          cnt = ;
  55.          root = newnode();
  56.      }
  57.  
  58.      void insert(char buf[]) {
  59.          int len = strlen(buf);
  60.          int now = root;
  61.          for (int i = ; i < len; i++) {
  62.              if (next[now][buf[i] - 'a'] == -) next[now][buf[i] - 'a'] = newnode();
  63.              now = next[now][buf[i] - 'a'];
  64.          }
  65.          End[now]++;
  66.      }
  67.  
  68.      void build() {
  69.          queue<int> Q;
  70.          fail[root] = root;
  71.          for (int i = ; i < ; i++)
  72.              if (next[root][i] == -) next[root][i] = root;
  73.              else {
  74.                  fail[next[root][i]] = root;
  75.                  Q.push(next[root][i]);
  76.              }
  77.          while (!Q.empty()) {
  78.              int now = Q.front();
  79.              Q.pop();
  80.              for (int i = ; i < ; i++)
  81.                  if (next[now][i] == -) next[now][i] = next[fail[now]][i];
  82.                  else {
  83.                      fail[next[now][i]] = next[fail[now]][i];
  84.                      Q.push(next[now][i]);
  85.                  }
  86.          }
  87.      }
  88.  
  89.      int query(char buf[]) {
  90.          int len = strlen(buf);
  91.          int now = root;
  92.          int res = ;
  93.          for (int i = ; i < len; i++) {
  94.              now = next[now][buf[i] - 'a'];
  95.              int temp = now;
  96.              while (temp != root) {
  97.                  res += End[temp];
  98.                  End[temp] = ;
  99.                  temp = fail[temp];
  100.              }
  101.          }
  102.          return res;
  103.      }
  104.  
  105.      void debug() {
  106.          for (int i = ; i < cnt; i++) {
  107.              printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
  108.              for (int j = ; j < ; j++) printf("%2d", next[i][j]);
  109.              printf("]\n");
  110.          }
  111.      }
  112.  }ac;
  113.  
  114.  char buf[];
  115.  int T,n;
  116.  int main() {
  117.      sf(T);
  118.      while(T--){
  119.          sf(n);
  120.          ac.init();
  121.          for (int i = ; i < n; ++i) {
  122.              scanf("%s",buf);
  123.              ac.insert(buf);
  124.          }
  125.          scanf("%s",buf);
  126.          ac.build();
  127.          printf("%d\n",ac.query(buf));
  128.      }
  129.      return ;
  130.  }

病毒侵袭 HDU - 2896

给出n个模式串，对m个串进行匹配，输出匹配的模式串的编号，以及总共多少个串可以匹配。

字符总数有128，而且编号要排序。

  1.  #include <set>
  2.  #include <map>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <cstdio>
  7.  #include <string>
  8.  #include <vector>
  9.  #include <time.h>
  10.  #include <cstring>
  11.  #include <iostream>
  12.  #include <algorithm>
  13.  #include <unordered_map>
  14.  
  15.  #define  pi acos(-1.0)
  16.  #define  eps 1e-9
  17.  #define  fi first
  18.  #define  se second
  19.  #define  rtl   rt<<1
  20.  #define  rtr   rt<<1|1
  21.  #define  bug               printf("******\n")
  22.  #define  mem(a, b)         memset(a,b,sizeof(a))
  23.  #define  name2str(x)       #x
  24.  #define  fuck(x)           cout<<#x" = "<<x<<endl
  25.  #define  sf(n)             scanf("%d", &n)
  26.  #define  sff(a, b)         scanf("%d %d", &a, &b)
  27.  #define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
  28.  #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
  29.  #define  pf                printf
  30.  #define  FIN               freopen("../date.txt","r",stdin)
  31.  #define  gcd(a, b)         __gcd(a,b)
  32.  #define  lowbit(x)         x&-x
  33.  #define  IO                iOS::sync_with_stdio(false)
  34.  
  35.  using namespace std;
  36.  typedef long long LL;
  37.  typedef unsigned long long ULL;
  38.  const int maxn = 1e6 + ;
  39.  const int maxm = 8e6 + ;
  40.  const int INF = 0x3f3f3f3f;
  41.  const int mod = 1e9 + ;
  42.  
  43.  struct Aho_Corasick {
  44.      int next[][], fail[], End[],vis[];
  45.      int root, cnt;
  46.      vector<int>ans;
  47.      int newnode() {
  48.          for (int i = ; i < ; i++) next[cnt][i] = -;
  49.          End[cnt++] = ;
  50.          return cnt - ;
  51.      }
  52.  
  53.      void init() {
  54.          cnt = ;
  55.          root = newnode();
  56.      }
  57.  
  58.      void insert(char buf[],int id) {
  59.          int len = strlen(buf);
  60.          int now = root;
  61.          for (int i = ; i < len; i++) {
  62.              if (next[now][buf[i]] == -) next[now][buf[i]] = newnode();
  63.              now = next[now][buf[i]];
  64.          }
  65.          End[now]++;
  66.          vis[now]=id;
  67.      }
  68.  
  69.      void build() {
  70.          queue<int> Q;
  71.          fail[root] = root;
  72.          for (int i = ; i < ; i++)
  73.              if (next[root][i] == -) next[root][i] = root;
  74.              else {
  75.                  fail[next[root][i]] = root;
  76.                  Q.push(next[root][i]);
  77.              }
  78.          while (!Q.empty()) {
  79.              int now = Q.front();
  80.              Q.pop();
  81.              for (int i = ; i < ; i++)
  82.                  if (next[now][i] == -) next[now][i] = next[fail[now]][i];
  83.                  else {
  84.                      fail[next[now][i]] = next[fail[now]][i];
  85.                      Q.push(next[now][i]);
  86.                  }
  87.          }
  88.      }
  89.  
  90.      int query(char buf[]) {
  91.          int len = strlen(buf);
  92.          int now = root;
  93.          int res = ;
  94.          for (int i = ; i < len; i++) {
  95.              now = next[now][buf[i]];
  96.              int temp = now;
  97.              while (temp != root) {
  98.                  res += End[temp];
  99.                  if (vis[temp]) ans.push_back(vis[temp]);
  100.  //                End[temp] = 0;
  101.                  temp = fail[temp];
  102.              }
  103.          }
  104.          return res;
  105.      }
  106.      void pf_ans(){
  107.          sort(ans.begin(),ans.end());
  108.          for (int i= ;i<ans.size() ;i++) printf(" %d",ans[i]);
  109.          printf("\n");
  110.          ans.clear();
  111.      }
  112.      void debug() {
  113.          for (int i = ; i < cnt; i++) {
  114.              printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
  115.              for (int j = ; j < ; j++) printf("%2d", next[i][j]);
  116.              printf("]\n");
  117.          }
  118.      }
  119.  }ac;
  120.  
  121.  char buf[];
  122.  int n,m;
  123.  int main() {
  124.     // FIN;
  125.      while(~sf(n)){
  126.          ac.init();
  127.          for (int i = ; i <= n; ++i) {
  128.              scanf("%s",buf);
  129.              ac.insert(buf,i);
  130.          }
  131.          ac.build();
  132.          sf(m);
  133.          int tot=;
  134.          for (int i = ; i <= m; ++i) {
  135.              scanf("%s",buf);
  136.              //fuck(i);
  137.              if (ac.query(buf)) {
  138.                  tot++;
  139.                  printf("web %d:",i);
  140.                  ac.pf_ans();
  141.              }
  142.          }
  143.          printf("total: %d\n",tot);
  144.      }
  145.      return ;
  146.  }

病毒侵袭持续中 HDU - 3065

这题题意和上一题一样，多了一个输出出现次数。

  1.  #include <set>
  2.  #include <map>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <cstdio>
  7.  #include <string>
  8.  #include <vector>
  9.  #include <time.h>
  10.  #include <cstring>
  11.  #include <iostream>
  12.  #include <algorithm>
  13.  #include <unordered_map>
  14.  
  15.  #define  pi acos(-1.0)
  16.  #define  eps 1e-9
  17.  #define  fi first
  18.  #define  se second
  19.  #define  rtl   rt<<1
  20.  #define  rtr   rt<<1|1
  21.  #define  bug               printf("******\n")
  22.  #define  mem(a, b)         memset(a,b,sizeof(a))
  23.  #define  name2str(x)       #x
  24.  #define  fuck(x)           cout<<#x" = "<<x<<endl
  25.  #define  sf(n)             scanf("%d", &n)
  26.  #define  sff(a, b)         scanf("%d %d", &a, &b)
  27.  #define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
  28.  #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
  29.  #define  pf                printf
  30.  #define  FIN               freopen("../date.txt","r",stdin)
  31.  #define  gcd(a, b)         __gcd(a,b)
  32.  #define  lowbit(x)         x&-x
  33.  #define  IO                iOS::sync_with_stdio(false)
  34.  
  35.  using namespace std;
  36.  typedef long long LL;
  37.  typedef unsigned long long ULL;
  38.  const int maxn = 1e6 + ;
  39.  const int maxm = 8e6 + ;
  40.  const int INF = 0x3f3f3f3f;
  41.  const int mod = 1e9 + ;
  42.  
  43.  char str[][];
  44.  char buf[];
  45.  int n, m;
  46.  struct Aho_Corasick {
  47.      int next[*][], fail[*], End[*], num[];
  48.      int root, cnt;
  49.  
  50.      int newnode() {
  51.          for (int i = ; i < ; i++) next[cnt][i] = -;
  52.          End[cnt++] = ;
  53.          return cnt - ;
  54.      }
  55.  
  56.      void init() {
  57.          cnt = ;
  58.          root = newnode();
  59.      }
  60.  
  61.      void insert(char buf[], int id) {
  62.          int len = strlen(buf);
  63.          int now = root;
  64.          for (int i = ; i < len; i++) {
  65.              if (next[now][buf[i]] == -) next[now][buf[i]] = newnode();
  66.              now = next[now][buf[i]];
  67.          }
  68.          End[now] = id;
  69.      }
  70.  
  71.      void build() {
  72.          queue<int> Q;
  73.          fail[root] = root;
  74.          for (int i = ; i < ; i++)
  75.              if (next[root][i] == -) next[root][i] = root;
  76.              else {
  77.                  fail[next[root][i]] = root;
  78.                  Q.push(next[root][i]);
  79.              }
  80.          while (!Q.empty()) {
  81.              int now = Q.front();
  82.              Q.pop();
  83.              for (int i = ; i < ; i++)
  84.                  if (next[now][i] == -) next[now][i] = next[fail[now]][i];
  85.                  else {
  86.                      fail[next[now][i]] = next[fail[now]][i];
  87.                      Q.push(next[now][i]);
  88.                  }
  89.          }
  90.      }
  91.  
  92.      void query(char buf[]) {
  93.          for(int i = ;i <= n;i++) num[i] = ;
  94.          int len = strlen(buf);
  95.          int now = root;
  96.          for (int i = ; i < len; i++) {
  97.              now = next[now][buf[i]];
  98.              int temp = now;
  99.              while (temp != root) {
  100.                if (End[temp])  num[End[temp]]++;
  101.  //                End[temp] = 0;
  102.                  temp = fail[temp];
  103.              }
  104.          }
  105.          for (int i= ;i<=n ;i++)
  106.            if (num[i])  printf("%s: %d\n",str[i],num[i]);
  107.      }
  108.  
  109.      void debug() {
  110.          for (int i = ; i < cnt; i++) {
  111.              printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
  112.              for (int j = ; j < ; j++) printf("%2d", next[i][j]);
  113.              printf("]\n");
  114.          }
  115.      }
  116.  } ac;
  117.  
  118.  int main() {
  119.      //FIN;
  120.      while (~sf(n)) {
  121.          ac.init();
  122.          for (int i = ; i <= n; ++i) {
  123.              scanf("%s", str[i]);
  124.              ac.insert(str[i], i);
  125.          }
  126.          ac.build();
  127.          scanf("%s", buf);
  128.          ac.query(buf);
  129.      }
  130.      return ;
  131.  }

Detect the Virus ZOJ - 3430

这题本质上还是和上面一个套路只是恶心了很多。

题目是给先你n组经过编码后的病毒串：

编码就是将原串转成二进制后取每6位进行转化成10进制后与上面表格对应的字符串，

末尾不足6位补0，且原串个数为3k+1则在编码后的串里增加'==',

若个数为3k+2则增加'=',给定的模式串也是编码后的串，

所以需要进行反编码然后就是赤裸裸的AC自动机模版了。

注意转化后的字符范围为256（包括转义字符），strlen也无法使用，需要用unsigned char

  1.  #include <cstdio>
  2.  #include <cstring>
  3.  #include <queue>
  4.  
  5.  #define  sf(n)             scanf("%d", &n)
  6.  #define  sff(a, b)         scanf("%d %d", &a, &b)
  7.  
  8.  using namespace std;
  9.  typedef long long LL;
  10.  typedef unsigned long long ULL;
  11.  const int maxn = 1e6 + ;
  12.  const int maxm = 8e6 + ;
  13.  const int INF = 0x3f3f3f3f;
  14.  const int mod = 1e9 + ;
  15.  
  16.  unsigned char buf[];
  17.  char str[];
  18.  unsigned char s[];
  19.  int n, m, tot;
  20.  
  21.  struct Aho_Corasick {
  22.      int next[ * ][], fail[ * ], End[ * ], vis[];
  23.      int root, cnt;
  24.  
  25.      int newnode() {
  26.          for (int i = ; i < ; i++) next[cnt][i] = -;
  27.          End[cnt++] = ;
  28.          return cnt - ;
  29.      }
  30.  
  31.      void init() {
  32.          cnt = ;
  33.          root = newnode();
  34.      }
  35.  
  36.      void insert(unsigned char buf[], int len, int id) {
  37.          int now = root;
  38.          for (int i = ; i < len; i++) {
  39.              if (next[now][buf[i]] == -) next[now][buf[i]] = newnode();
  40.              now = next[now][buf[i]];
  41.          }
  42.          End[now] = id;
  43.      }
  44.  
  45.      void build() {
  46.          queue<int> Q;
  47.          fail[root] = root;
  48.          for (int i = ; i < ; i++)
  49.              if (next[root][i] == -) next[root][i] = root;
  50.              else {
  51.                  fail[next[root][i]] = root;
  52.                  Q.push(next[root][i]);
  53.              }
  54.          while (!Q.empty()) {
  55.              int now = Q.front();
  56.              Q.pop();
  57.              for (int i = ; i < ; i++)
  58.                  if (next[now][i] == -) next[now][i] = next[fail[now]][i];
  59.                  else {
  60.                      fail[next[now][i]] = next[fail[now]][i];
  61.                      Q.push(next[now][i]);
  62.                  }
  63.          }
  64.      }
  65.  
  66.      int query(unsigned char buf[], int len, int n) {
  67.          memset(vis, false, sizeof(vis));
  68.          int now = root;
  69.          for (int i = ; i < len; i++) {
  70.              now = next[now][buf[i]];
  71.              int temp = now;
  72.              while (temp != root) {
  73.                  if (End[temp] != -)
  74.                      vis[End[temp]] = true;
  75.                  temp = fail[temp];
  76.              }
  77.          }
  78.          int res = ;
  79.          for (int i = ; i <= n; i++) if (vis[i]) res++;
  80.          return res;
  81.      }
  82.  
  83.      void debug() {
  84.          for (int i = ; i < cnt; i++) {
  85.              printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
  86.              for (int j = ; j < ; j++) printf("%2d", next[i][j]);
  87.              printf("]\n");
  88.          }
  89.      }
  90.  } ac;
  91.  
  92.  unsigned char Get(char ch) {
  93.      if (ch >= 'A' && ch <= 'Z')return ch - 'A';
  94.      if (ch >= 'a' && ch <= 'z')return ch - 'a' + ;
  95.      if (ch >= '' && ch <= '')return ch - '' + ;
  96.      if (ch == '+')return ;
  97.      else return ;
  98.  }
  99.  
  100.  void change(unsigned char str[], int len) {
  101.      int t = ;
  102.      for (int i = ; i < len; i += ) {
  103.          buf[t++] = ((str[i] << ) | (str[i + ] >> ));
  104.          if (i +  < len)
  105.              buf[t++] = ((str[i + ] << ) | (str[i + ] >> ));
  106.          if (i +  < len)
  107.              buf[t++] = ((str[i + ] << ) | str[i + ]);
  108.      }
  109.      tot = t;
  110.  }
  111.  
  112.  int main() {
  113.      while (~sf(n)) {
  114.          ac.init();
  115.          for (int i = ; i <= n; ++i) {
  116.              scanf("%s", str);
  117.              int len = strlen(str);
  118.              while (str[len - ] == '=')len--;
  119.              for (int j = ; j < len; j++) s[j] = Get(str[j]);
  120.              change(s, len);
  121.              ac.insert(buf, tot, i);
  122.          }
  123.          ac.build();
  124.          sf(m);
  125.          for (int i = ; i <= m; i++) {
  126.              scanf("%s", str);
  127.              int len = strlen(str);
  128.              while (str[len - ] == '=') len--;
  129.              for (int j = ; j < len; j++) s[j] = Get(str[j]);
  130.              change(s, len);
  131.              printf("%d\n", ac.query(buf, tot, n));
  132.          }
  133.          printf("\n");
  134.      }
  135.      return ;
  136.  }

HDU - 2222,HDU - 2896,HDU - 3065,ZOJ - 3430 AC自动机求文本串和模式串信息(模板题)的更多相关文章

  1. Detect the Virus ZOJ - 3430 AC自动机

    One day, Nobita found that his computer is extremely slow. After several hours' work, he finally fou ...

  2. ZOJ - 3430 ac自动机

    这题主要就是解码过程很恶心,不能用char存,一共wa了20发 题意:先给n串加密后的字符,然后m串加密后的字符,解码之后求n对应每个m的匹配数,很显然的ac自动机 加密过程是先用对应ascii表的标 ...

  3. hdu 3695:Computer Virus on Planet Pandora（AC自动机，入门题）

    Computer Virus on Planet Pandora Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 256000/1280 ...

  4. HDU 2243 考研路茫茫——单词情结（AC自动机+矩阵）

    考研路茫茫——单词情结 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  5. HDU 2846 (AC自动机+多文本匹配)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2846 题目大意:有多个文本,多个模式串.问每个模式串中,有多少个文本?(匹配可重复) 解题思路: 传统 ...

  6. HDU 2243 考研路茫茫——单词情结（AC自动机+矩阵快速幂）

    http://acm.hdu.edu.cn/showproblem.php?pid=2243 题意: 给出m个模式串,求长度不超过n的且至少包含一个模式串的字符串个数. 思路: 如果做过poj2778 ...

  7. HDU 6138 Fleet of the Eternal Throne（AC自动机）

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6138 [题目大意] 给出一些串,询问第x个串和第y个串的公共子串, 同时要求该公共子串为某个串的前 ...

  8. HDU 2896：病毒侵袭（AC自动机）

    http://acm.hdu.edu.cn/showproblem.php?pid=2896 题意:中文题意. 思路:AC自动机模板题.主要在于字符有128种,输出还要排序和去重! 注意是“total ...

  9. HDU 3695 Computer Virus on Planet Pandora（AC自动机模版题）

    Computer Virus on Planet Pandora Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 256000/1280 ...

随机推荐

  1. Jmeter-【JSON Extractor】-响应结果中三级key取值

    一.请求返回样式 二.取第三个option 三.查看结果

  2. delphi根据不同图片生成不规则窗口的实现(仅限于BMP格式)

    unit CreateImageForm; interface uses Windows, SysUtils, Variants, Classes, Graphics; procedure Creat ...

  3. NX二次开发-UFUN求两个对象最短距离UF_MODL_ask_minimum_dist

    NX9+VS2012 #include <uf.h> #include <uf_modl.h> #include <uf_ui.h> UF_initialize() ...

  4. NX二次开发-UFUN结合NXOPEN开发_常用代码模板

    hpp //头文件 #include <NXOpen/Part.hxx> #include <NXOpen/PartCollection.hxx> #include <N ...

  5. flutter 插件

    flutter_spinkit loading动画

  6. iOS开发之UIMenuController

    1.简介 用于剪切.复制.粘贴.选择.选择ALL和DELETE命令的菜单界面. UITextField.UIWebView.UITextView自带有这种UIMenuController效果: 2.相 ...

  7. Debian怎么配置网卡(IP)

    来自:http://jingyan.baidu.com/article/a3f121e4d27a53fc9152bb65.html Debian可以配置静态IP.动态IP使Debian连上互联网.用户 ...

  8. 2019 USP Try-outs 练习赛

    // 好久没更博客了,最近打了很多场练习赛&校内PK赛,大概自闭忙于补题吧 // 9.26 周四练习赛 A. Kolkhozy 题意 有 n 个数 \(f[i]\) ,有 q 次询问(l, r ...

  9. Redis连不上的一些细节配置

    远程链接redis连不上,在确保防火墙设置正确的情况下 把redis.conf中的 bind 127.0.0.1注释 另外把protected-mode yes 改为protected-mode no

  10. 解决WIN8输入法的问题，Ctrl+空格，Ctrl+Shift，切换问题

    在WIN8中,我们曾经熟悉的的Ctrl+空格和Ctrl+Shift消失了,取而导致的是WIN+空格. 在这里先简单解释一下WIN8的输入法结构: 在WIN7以前的输入法中,输入法采用了平行目录的结构, ...