NY-5

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

30
3 

/*字符串的匹配问题，通过对字符串的每四位截取然后与匹配串进行匹配，如果相等strcmp==0则count++,这样的问题是RUNTIME，解决办法是通过还没到四位之前只要判断有一位与对应的匹配串不同则直接进行到下一位的匹配 */#include<stdio.h>#include<string.h>int main(){int n,k=0;char A[15],B[1010],s[15];scanf("%d",&n);while(n--){scanf("%s",A);scanf("%s",B);int lenb=strlen(B);int lena=strlen(A);int count=0;int j=0;for(int i=0;i<lenb;i++){s[k++]=B[i];s[k]='\0';if(k==lena){if(strcmp(s,A)==0)count++;i=j;j++;k=0;}else{if(s[k-1]!=A[k-1]){i=j;j++;k=0;}}}printf("%d\n",count);}return 0;}