题目5：Binary String Matching

题目链接：

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

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

算法思想：

可以使用普通的循环遍历加回溯就可以，因为目标串最大10000个字符，模式串最大10个字符，故不会超时。当然也可以使用KMP算法。还有更简单的就是使用C++STL库中string类的find的函数。

源代码

/*
Author:杨林峰
Date:2017.10.20
NYOJ(5):Binary String Matching
*/
#include <iostream>
#include <string>
using namespace std;
int getNum(string pattern, string target)
{
int len1 = pattern.length(), len2 = target.length();
int i, j, ans = 0;
for (i = 0; i <= len2 - len1; i++)
{
int tmp = i;
for (j = 0; j < len1;)
{
//当模式串与目标串字符相等时，继续比较下一个字符
if (pattern[j] == target[i])
{
i++;
j++;
}
//否则，跳出循环
else
{
break;
}
//cout << i << " " << j << endl;
}
//判断目标串是否成功
if (j >= len1)
ans++;
//回溯
i = tmp;
j = 0;
}
return ans;
}
int main()
{
int N, ans;
string pattern, target;
cin >> N;
while (N--)
{
cin >> pattern >> target;
ans = getNum(pattern, target);
cout << ans << endl;
}
return 0;
}

最优源代码

#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;
}
}