cf-404A Valera and X

最新推荐文章于 2024-10-27 21:04:33 发布

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最新推荐文章于 2024-10-27 21:04:33 发布

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文章标签： cf

本文链接：https://blog.youkuaiyun.com/Floyld/article/details/52792095

本文介绍了一道编程题，任务是判断一个n×n的字符矩阵是否构成字母“X”。通过对角线上的字符和非对角线上的字符进行比较来验证。文章提供了完整的代码实现，包括输入输出和核心逻辑。

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Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals n squares (n is an odd number) and each unit square contains some small letter of the English alphabet.

Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:

on both diagonals of the square paper all letters are the same;
all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.

Help Valera, write the program that completes the described task for him.

Input

The first line contains integer n (3 ≤ n < 300; n is odd). Each of the next n lines contains n small English letters — the description of Valera's paper.

Output

Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.

Examples

input
5
xooox
oxoxo
soxoo
oxoxo
xooox

output
NO

input
3
wsw
sws
wsw

output
YES

input
3
xpx
pxp
xpe

output
NO
题目分析：判断给出的字符矩阵的对角线是否是全都一样的字符，除了对角线外的字符是否全都是一样的。    ！！！！注意：对角线与除对角线外的字符是不能相同的。#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

char a[310][310];

int main()
{
    int n;
    while (cin >> n)
    {
        for (int i = 0; i < n; i++)
            cin >> a[i];
        int f = 1;
        char c = a[0][0], d = a[0][1];
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if ((i == j) || (i + j == n - 1))
                {
                    if (a[i][j] != a[0][0])
                    {
                        f = 0;
                        break;
                    }
                }
                else
                {
                    if (a[i][j] != a[0][1])
                    {
                        f = 0;
                        break;
                    }
                }
            }
        }
        if (f == 1)
        {
            if (c == d)
                cout << "NO" << endl;
            else
                cout << "YES" << endl;
        }
        else
            cout << "NO" << endl;
    }
    return 0;
}