Binary String Matching
时间限制:3000 ms | 内存限制:65535 KB
难度:3
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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
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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.
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样例输入
-
3
11
1001110110
101
110010010010001
1010
110100010101011
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样例输出
-
3
0
3
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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;
}
NYOJ 5 Binary String Matching【string find的运用】
原文:http://blog.csdn.net/hurmishine/article/details/52261820