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?$T$?($0<T\le 100$) , indicating the cases following. In the next T lines each line contains two strings —?$A$?and?$B$. (?$0<|A|\le {10}^5,0<|B|\le {10}^5$)

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.

題意：問給你兩個字符串A和B，其中A的后綴也是B的前綴的子串有多少個。

題解：很簡單的字符串hash問題，

設i是從 0 循環到min(lenA,lenB)

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

hashb = hashb * Base + B[i];

然后判斷hasha == hashb 那么cnt++

代碼：

#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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