南阳oj 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

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

#include<stdio.h>
#include<string.h>
int main()
{
    char a[120],b[1200];
    int ai,bi;
    int t,i,j,k,cut;
    scanf("%d",&t);
    while(t--){
        cut=0;
        scanf("%s%s",&a,b);
        ai=a[0]-'0';
        int len1=strlen(a);
        int len2=strlen(b);
        for(i=1;i<len1;i++){
            ai*=10;
            ai+=a[i]-'0';
        }
        for(i=0;i<len2;i++)
        {
            bi=b[i]-'0';
            for(k=i+1,j=0;j<len1-1;j++,k++)
            {
                bi*=10;
                bi+=b[k]-'0';
            }
            if(ai==bi)   cut++;
        }
        printf("%d
",cut);
    }
    return 0;
}