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("*********
");
#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-8;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
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[1] = 0;
    for(int i = 2, j = 0; i <= lens2; i++) {
        while(j > 0 && s2[i] != s2[j+1]) j = nex[j];
        if(s2[i] == s2[j+1]) j++;
        nex[i] = j;
    }
}

void kmp() {
    get_next();
    for(int i = 1, j = 0; i <= lens1; i++) {
        while(j > 0 && (j == lens2 || s1[i] != s2[j+1])) j = nex[j];
        if(s1[i] == s2[j+1]) j++;
        if(j == lens2) {
            num++;
        }
    }
}

bool check(int x) {
    lens2 = x;
    num = 0;
    kmp();
    return num >= k;
}

int main() {
    fgets(s1 + 1, maxn, stdin);
    fgets(s2 + 1, maxn, stdin);
    scanf("%d", &k);
    lens1 = strlen(s1 + 1), lens2 = strlen(s2 + 1);
    int ub = lens2, lb = 0, mid;
    ans = 0;
    while(ub >= lb) {
        mid = (ub + lb) >> 1;
        if(check(mid)) {
            ans = mid;
            lb = mid + 1;
        } else {
            ub = mid - 1;
        }
    }
    if(ans > 0) {
        for(int i = 1; i <= ans; i++) {
            printf("%c", s2[i]);
        }
        printf("
");
    }
    else printf("IMPOSSIBLE
");
    return 0;
}