nyoj 5 Binary String Matching 【string】

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

STL string find函数：  ans=str.find(s.ans);[查找的偏移]

ans!=string::npos;[查找结束的意思]

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
int main()
{
int t;
string str,s;
cin>>t;
while(t--)
{
cin>>s>>str;//C++输入
int ans=0;
int cnt=0;
while((ans=str.find(s,ans))!=string::npos)
{
cnt++;
ans++;
}
cout<<cnt<<endl;
}
return 0;
}