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<iostream>
#include<string>
#include<stdio.h>
using namespace std;
int main()
{
int N;
scanf("%d",&N);
while(N--)
{
string a, b;
cin>>a>>b;
int count = 0;
int position = b.find(a,0);//find 函数 返回jk 在s 中的下标位置
while(position!=string::npos)
{
count++;
position = b.find(a,position+1);
}
printf("%d\n",count);
}
return 0;
}
/*先说说string::npos参数：
npos 是一个常数，用来表示不存在的位置，类型一般是std::container_type::size_type 许多容器都提供这个东西。取值由实现决定，一般是-1，这样做，就不会存在移植的问题了。
再来说说find函数：
find函数的返回值是整数，假如字符串存在包含关系，其返回值必定不等于npos，但如果字符串不存在包含关系，那么返回值就一定是npos。*/