Binary String Matching

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

上传者

naonao

循环AC

#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
using namespace std;

#define mem(a) memset(a, 0, sizeof(a))

char a[100], b[1100];

int main() {
int n;
cin >> n;
while (n --) {
mem(a); mem(b);
cin >> a >> b;
int res = 0, p1 = strlen(a), p2 = strlen(b);
int num = 0;
while (num <= p2-p1) {
int flag = 1;
for (int i = num, j = 0; j<p1; j++,i++) {
if (a[j] != b[i])   flag = 0;
}
if (flag)   res ++ ;
num ++ ;
//cout << res << endl;
}
cout << res << endl;
}
return 0;
}

标程：#include中的find()函数的应用。另外，m!=string::npos 意思是：m不等于字符串的尾部。

#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;
}
}