1087: Common Substrings （哈希）

1087: Common Substrings

Time Limit:3000/1000 MS (Java/Others)   Memory Limit:163840/131072 KB (Java/Others)
Total Submissions:857   Accepted:112

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Description

You are given two long strings A

and B. They are comprised of lowercase letters. You should compute how many suffixes of A are the prefixes of B

.

Input

In the first line is a number T

(0<T≤100

) , indicating the cases following.

In the next T lines each line contains two strings — A

and B

.

( 0<|A|≤105,0<|B|≤105

)

Output

There should be exactly T

lines.

Each line contain a number — the answer.

Sample Input

1
abcc ccba

Sample Output

2

HINT

In the first case, cc and c are two of the suffixes of string A, and they are the prefixes of string B.

【分析】H[i]-H[L]*xp[L]表示从s[i]开始的长度为L的字符串的哈希值。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
using namespace std;
typedef long long ll;
const ll N = 1e5+;
const int x=;
ll H1[N],H2[N],xp[N];
ll hs1,hs2;
int ans;
char stra[N],strb[N];
int main()
{
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%s%s",stra,strb);
        int lena=strlen(stra),lenb=strlen(strb);
        H1[lena]=,H2[lenb]=;
        for (int i=lena-;i>=;--i)
            H1[i]=H1[i+]*x+(stra[i]-'a');
        for (int i=lenb-;i>=;--i){
            H2[i]=H2[i+]*x+(strb[i]-'a');
        }
        xp[]=;
        ans=;
        for (int i=;i<=max(lena,lenb);++i)
            xp[i]=xp[i-]*x;
        for (int L=;L<=min(lena,lenb);++L){
            hs2=H2[]-H2[L]*xp[L];
            hs1=H1[lena-L]-H1[lena]*xp[L];
            if (hs2==hs1)
                ans++;
        }
        printf("%d\n",ans);
    }
    return ;
}