E.Text Editor （Gym 101466E + 二分 + kmp)

题目链接：http://codeforces.com/gym/101466/problem/E

题目：

题意：

　　给你s串和t串，一个数k，求t的最长前缀串在s串中出现次数不少于k。

思路：

　　一眼二分+kmp，二分最长前缀串的长度即可。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define lowbit(x) x&(-x)
 #define bug printf("*********\n");
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = 1e9 + ;
 const int maxn = 1e5 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 int k, lens1, lens2, ans, num;
 char s1[maxn], s2[maxn];
 int nex[maxn];

 void get_next() {
     nex[] = ;
     for(int i = , j = ; i <= lens2; i++) {
         while(j >  && s2[i] != s2[j+]) j = nex[j];
         if(s2[i] == s2[j+]) j++;
         nex[i] = j;
     }
 }

 void kmp() {
     get_next();
     for(int i = , j = ; i <= lens1; i++) {
         while(j >  && (j == lens2 || s1[i] != s2[j+])) j = nex[j];
         if(s1[i] == s2[j+]) j++;
         if(j == lens2) {
             num++;
         }
     }
 }

 bool check(int x) {
     lens2 = x;
     num = ;
     kmp();
     return num >= k;
 }

 int main() {
     fgets(s1 + , maxn, stdin);
     fgets(s2 + , maxn, stdin);
     scanf("%d", &k);
     lens1 = strlen(s1 + ), lens2 = strlen(s2 + );
     int ub = lens2, lb = , mid;
     ans = ;
     while(ub >= lb) {
         mid = (ub + lb) >> ;
         if(check(mid)) {
             ans = mid;
             lb = mid + ;
         } else {
             ub = mid - ;
         }
     }
     if(ans > ) {
         for(int i = ; i <= ans; i++) {
             printf("%c", s2[i]);
         }
         printf("\n");
     }
     else printf("IMPOSSIBLE\n");
     return ;
 }