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  • (find) nyoj5-Binary String Matching

    5-Binary String Matching


    内存限制:64MB 时间限制:3000ms 特判: No
    通过数:232 提交数:458 难度:3

    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:

    3
    11
    1001110110
    101
    110010010010001
    1010
    110100010101011 

    样例输出:

    3
    0
    3 

    可以用find。
    注意用getchar()。。。。。。。
    C++代码:
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<cstdio>
    using namespace std;
    int main(){
        int T;
        scanf("%d",&T);
        getchar();  //必须加上
        while(T--){
            string s1;
            string s2;
            getline(cin,s1);
            getline(cin,s2);
            int ans = s2.find(s1,0);
            int num = 0;
            while(ans >= 0){
                num++;
                ans = s2.find(s1,ans+1);
            }
            printf("%d
    ",num);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Weixu-Liu/p/10532078.html
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