公共子串字符串哈希

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 \leq 100

) , indicating the cases following.

In the next T lines each line contains two strings —

A

and

B

(

0 < | A | \leq 10^{5}, 0 < | B | \leq 10^{5}

)

Output

There should be exactly

T

lines.

Each line contain a number — the answer.

Sample Input

1
abcc ccba

Sample Output

HINT

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

题意：问给你两个字符串A和B，其中A的后缀也是B的前缀的子串有多少个。

题解：很简单的字符串hash问题，

设i是从 0 循环到min(lenA,lenB)

hasha += Base^n * A[lenA-1-i];

hashb = hashb * Base + B[i];

然后判断hasha == hashb 那么cnt++

代码：

#include <cstdio>
#include <iostream>
#include <cstring> 
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
const ull Base = 100000007;
const ull MAX = 1e5+5;
char A[MAX],B[MAX];
int main()
{int T;scanf("%lld",&T);while(T--){scanf("%s %s",A,B);ull lenA = strlen(A),ull asum = 0;ull lenB = strlen(B),ull bsum = 0;ull t = 1;int ans = 0;for(int  i = 0;i < min(lenA,lenB);i++){asum += t*A[lenA-1-i];bsum = bsum * Base + B[i];t *= Base;if(asum == bsum) ans++;}cout<<ans<<endl;}return 0;
}

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