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5-Binary String Matching

內存限制:64MB 時間限制:3000ms 特判: No
通過數:232 提交數:458 難度:3

題目描述:

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

輸入描述:

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.

輸出描述:

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:

復制

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出:

3
0
3 

可以用find。
註意用getchar()。。。。。。。
C++代碼：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
int main(){
    int T;
    scanf("%d",&T);
    getchar();  
 
//必須加上
    while(T--){
        string s1;
        string s2;
        getline(cin,s1);
        getline(cin,s2);
        int ans = s2.find(s1,0);
        int num = 0;
        while(ans >= 0){
            num++;
            ans = s2.find(s1,ans+1);
        }
        printf("%d\n",num);
    }
    return 0;
}

(find) nyoj5-Binary String Matching