CodeForces - 404A: Valera and X

最新推荐文章于 2021-02-15 18:37:11 发布

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本文介绍了一道编程题目，任务是判断一个n×n的字符矩阵是否构成字母X的形状。这里的X形状要求两个对角线上必须是相同的字符，而其他位置则为另一种统一的不同字符。文章提供了一个C语言实现的解决方案。

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

大致题意是对角线要同一种字符，其余位置是另外一种字符。

# include <stdio.h>
# include <string.h>
int main()
{
    int n, i, j ,flag = 0;
    char s[302][302], c, d;
    scanf("%d",&n);
    for(i=1; i<=n; ++i)
        scanf("%s",s[i]+1);
    c = s[1][1];//保存第一个字符
    d = s[1][2];//保存第二个字符
    if(c == d)
    {
        puts("NO");
        return 0;
    }
    for(i=1; i<=n/2; ++i)//先遍历到中间行
    {
        if(s[i][i] != s[i][n-i+1] || s[i][i] != c || s[i][i]==d)
            flag = 1;
        for(j=1; j<=n; ++j)
        {
            if(j==i || j==n-i+1)
                continue;
            else if(s[i][j] != d)
            {
                flag = 1;
                break;
            }
        }
        if(flag)
            break;
    }
    for(i=n/2+1; i<=n; ++i)//再遍历后面的行
    {
        if(s[i][n-i+1] != s[i][i] || s[i][i] != c || s[i][i]==d)
            flag = 1;
        for(j=1; j<=n; ++j)
        {
            if(j==n-i+1 || j==i)
                continue;
            else if(s[i][j] != d)
            {
                flag = 1;
                break;
            }
        }
        if(flag)
            break;
    }
    if(flag)
        puts("NO");
    else
        puts("YES");
    return 0;
}