Anti-Rhyme Pairs （求最长公共前缀）

题目链接：https://vjudge.net/contest/344930#problem/F

题目大意：给你n个字符串，让你求给定的两个串的最长公共前缀

题目思路：处理所给的n个字符串的Hash值，然后对于每次给定的两个串，二分长度就可以了。

值得注意的是这道题需要利用vector进行存储

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>


#define INF 0x3f3f3f3f
#define LL long long

typedef unsigned long long ull;
const int maxn = 1e5+10;

char s[maxn];
ull base = 131;
ull mod = 1e9+7;
ull p[maxn];
ull h1[maxn],h2[maxn];
ull q[maxn];
int len[maxn];
std::vector<ull> h[maxn];


ull get_hash(ull h[],int l,int r){
    return (h[r] - h[l-1]*p[r-l+1]);
}


int main() {
    int T;
    int cas = 1;
    scanf("%d",&T);
    while (T--) {
        int n;
        scanf("%d",&n);
        for (int i=1;i<=n;i++) {
            scanf("%s",s+1);
            len[i] = strlen(s+1);
            h[i].clear();
            h[i].push_back(0);
            for (int j=1;j<=len[i];j++) {
                h[i].push_back(h[i][j-1] * base + s[j] - 'a');
            }
        }
        int q;
        printf("Case %d:
",cas++);
        scanf("%d",&q);
        while (q--) {
            int u,v;
            scanf("%d%d",&u,&v);
            int l = 1,r = std::min(len[u],len[v]);
            int ans = 0;
            while (l <= r) {
                int m = (l + r) >> 1;
                if (h[u][m] == h[v][m]) {
                    ans = m;
                    l = m + 1;
                }
                else
                    r = m - 1;
            }
            printf("%d
",ans);
        }
    }
    return 0;
}