1411012118-ny-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 

题目大意

       给定两个字符串，求出字符串a在字符串b中出现的次数。

解题思路

       通过i++依次判断，注意对字符串的初始化。

代码

#include<stdio.h>#include<string.h>char a[12],b[1200],c[12];int main(){    int n;int i,j,k,sum;int len1,len2;scanf("%d",&n);getchar();while(n--){    scanf("%s%s",a,b);len1=strlen(a);len2=strlen(b);memset(c,0,sizeof(c));//每次都需要初始化，避免这次数据影响下次结果 sum=0;for(i=0;i<len2-len1+1;i++)//for(i=0;i<len2-len1;i++)    测试数据 111 1111 {    for(j=i,k=0;j<i+len1;j++,k++)        c[k]=b[j];    if(strcmp(a,c)==0)        sum++;}printf("%d\n",sum);}return 0;}