NYOJ-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

AC

#include <stdio.h>
#include <string.h> 
int main() 
{
    char a[10],b[1000];
    int N,m,n,i,j,sum;
    scanf("%d",&N);
    while(N--)
    {
        scanf("%s\n%s",a,b);//两个字符串输入
        m=strlen(a);n=strlen(b);
        sum=0;i=0;j=0;
        while(i<=n)
        {
            if(a[j]=='\0')
            {sum++;i=i-j+1;j=0;}//a[j]=='\0'时说明有一个与a相同,sum++
            else if(a[j]==b[i])
            {i++;j++;}//判断
            else
            {i=i-j+1;j=0;}//回到起点，再次从头判断
        }
        printf("%d\n",sum);
    }
    return 0;
}

回溯基本思想是：从问题的某一种状态（初始状态）出发，搜索从这种状态出发所能达到的所有“状态”，当一条路走到“尽头”的时候（不能再前进），再后退一步或若干步，从另一种可能“状态”出发，继续搜索，直到所有的“路径”（状态）都试探过。这种不断“前进”、不断“回溯”寻找解的方法，就称作“回溯法”。