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

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

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;
}
}