nyoj——5字符串STL复习

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

 

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

 

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

#include <bits/stdc++.h>
using namespace std;
typedef long long  ll;
const int INF = 0x3f3f3f3f;
const int maxn = 1000000;


int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        string pattern;
        string sym;
        cin >> pattern >>sym;
        int cnt = 0;
        while(sym.length()){
            if(sym.find(pattern) > 1005)
                break;
            int pos = sym.find(pattern);
            sym = sym.substr(pos+1);
            cnt++;
        }
        printf("%d
",cnt);
    }
    return 0;
}

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