NYOJ 5 Binary String Matching【string find的运用】

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 

来源

网络

原题链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=5

输入两个字符串，问你第二个字符串在第一个中出现了多少次，可重叠。

使用string 的find

#include <cstdio>#include <iostream>#include <cstring>using namespace std;string a,b;int main(){    int T;    //freopen("data/5.txt","r",stdin);    cin>>T;    while(T--)    {        cin>>a>>b;        int cnt=0;        //for (int index=0; (index=b.find(a,index))!=string::npos; index += 1,cnt++);        int index=0;        while((index=b.find(a,index))!=string::npos)        {            index++;            cnt++;        }        cout<<cnt<<endl;    }    return 0;}

OJ标程代码

 #include<iostream>#include<string>using namespace std;int main(){string s1,s2;int n;cin>>n;while(n--){cin>>s1>>s2;unsigned int m=s2.find(s1,0);int num=0;while(m!=string::npos){num++;m=s2.find(s1,m+1);}cout<<num<<endl;}}