nyoj_5 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

31110011101101011100100100100011010110100010101011 

样例输出

303 

分析：

不会KMP（o(╯□╰)o），好在数据比较水，只能暴力了。。。

代码：

#include<stdio.h>#include<string.h>int main(){int n;scanf("%d",&n);while(n--){char a[11],b[1001];scanf("%s",a);scanf("%s",b);int i,j,sum=0;for(j=0;j<strlen(b);j++){for(i=0;i<strlen(a);i++){if(b[j+i]!=a[i])break;}if(i==strlen(a))sum++;}printf("%d\n",sum);}return 0;}