NYOJ 5 Binary String Mathing (substr函数)

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

31110011101101011100100100100011010110100010101011 

样例输出

303 

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;int main(){int T;string a;string b;cin>>T;while(T--){cin>>a;cin>>b;int count=0;int len1=a.size();int len2=b.size();for(int i=0;i<len2;i++){        if(b.substr(i,len1)==a.substr(0,len1))        count++;}cout<<count<<endl;} return 0; }