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  • Codeforces Round #263 (Div. 2) proA

    称号:

    A. Appleman and Easy Task
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?

    Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.

    Input

    The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces.

    Output

    Print "YES" or "NO" (without the quotes) depending on the answer to the problem.

    Sample test(s)
    input
    3
    xxo
    xox
    oxx
    
    output
    YES
    
    input
    4
    xxxo
    xoxo
    oxox
    xxxx
    
    output
    NO

    题意分析:

    推断每一个位置相邻位置为o的数字,假设全为偶数就yes,否者no。简单模拟题。

    Orz


    代码:

    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <iostream>
    
    
    using namespace std;
    
    char a[105][105];
    int main()
    {
        int n;
        char c;
        int flag;
        cin >> n;
        memset(a, 0, sizeof(a));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                cin >> c;
                if (c == 'o')
                    a[i][j] = 1;
            }
        flag = 1;
    
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                int cnt = 0;
                if (a[i-1][j])
                    ++cnt;
                if (a[i+1][j])
                    ++cnt;
                if (a[i][j-1])
                    ++cnt;
                if (a[i][j+1])
                    ++cnt;
                if (cnt & 1)
                {
                    flag = false;
                    break;
                }
            }
        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    
        return 0;
    }


    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/4626051.html
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