nyoj 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

/**
    题目大意：
        求一个串的子串，再判断该串与题目所给的串是否相同
    分析：
        通过C++提供的substr函数求子串 
            substr用法：str2 = str1.substr (index, length)
**/

C/C++代码实现：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <stack>

using namespace std;

int N, len1, len2, cnt;

string s1, s2, s3;

int main () {
    scanf ("%d", &N);
    while (N --) {
        cnt = 0;
        cin >>s1 >>s2;
        len1 = s1.size();
        len2 = s2.size();
        for (int i = 0; i < len2 - len1 + 1; ++ i) {
            s3 = s2.substr(i, len1);
            if (s1 == s3) ++ cnt;
        } 
        printf ("%d\n", cnt);
    }
    return 0;
}

python实现（AC）：

 1 T = int(input())
 2 while T:
 3     T -= 1
 4     a = input()
 5     b = input()
 6     a_len = len(a)
 7     b_len = len(b)
 8     ans = 0
 9     for i in range(0, b_len - a_len + 1):
10         flag = 1
11         for j in range(0, a_len):
12             if b[i+j] != a[j]:
13                 flag = 0
14                 break
15         if flag:
16             ans += 1
17     print(ans)