acm--NSOJ-4617

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题目描述

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

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 
分享这道题，主要是分享C++中string.find()函数与string::npos的应用：
查找字符串a是否包含子串b,
strA.find(strB) != string:npos
string::size_type pos = strA.find(strB);
if(pos != string::npos){}
以此题举例说明：
#include<iostream>#include<algorithm>#include<cstdio>#include<string>using namespace std;int main(){    int t;    cin>>t;    while(t--) {        int i=0,sum=0;        string a,b;        cin>>a>>b;        while((i=b.find(a,i))!=(string::npos)){            i++; sum++;        }        cout<<sum<<endl;    }    return 0;}

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