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

輸入

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.

輸出

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

樣例輸入

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出

3
0
3

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main(int argc, char const *argv[]) {

    int N;

    scanf("%d",&N);

    while (N--) {
        //儲存 小字串
        char ch[10000];
        //儲存 大字串
        string str;
        int sum = 0;//記錄所含次數

        cin >> ch;
        cin >> str;

        //小字串長度
        int l = strlen(ch);

        //當str非空情況下
        while(!str.empty()){
            //返回大字串第一個元素的迭代器
            string::iterator ite = str.begin();
            //w 用於標誌
            int w = 0;
            //迴圈小字串
            for(int i = 0;i < l;i++){
                // 當不等於時標誌 且刪除第一個字串
                if(ch[i] != *ite){
                    w = 1;//標誌
                    str.erase(str.begin());
                    break;
                }
                ite++;
            }
            //標誌匹配時 匹配規模 +1 刪除第一個字串
            if(w == 0){
                sum++;
                str.erase(str.begin());
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}