5-Binary String Matching

原创于 2019-03-12 21:23:39 发布 · 134 阅读

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本文深入探讨了串匹配算法的基本原理，通过一个具体的编程实例，详细解释了如何在长字符串中查找特定子串出现的次数。文章提供了完整的代码实现，并附有输入输出样例，适合初学者理解串匹配的概念及其实现。

题目描述:
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
输入描述:
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.
输出描述:
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:
复制
3
11
1001110110
101
110010010010001
1010
110100010101011
样例输出:
3
0
3

这个题其实就是描述一个串里含有多少个另外一个串
不多说，上代码吧！

#include<stdio.h>
#include<string.h>
int main()
{
    int n;
    char a[11] , str[1010];
    while(~scanf("%d",&n))
    {
        getchar();
        while(n--)
        {
            gets(a);
            gets(str);
            int lena = strlen(a) ;
            int len = strlen(str);
            int sum = 0 ;
            int i = 0,j;
            for(int i=0; i<len; i++)
            {
                int k=i;
                for( j = 0 ; j < lena ; j++ ,k++)
                {
                    if(a[j] != str[k])
                        break;
                }
                if( j == lena )
                    sum++;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}