Magic Square

这道题是比较简单的，三个字——找规律。思路也应该是比较清晰的，下面，我们先看一下题目吧。
Description

In recreational mathematics, a magic square of n-degree is an arrangement of n2 numbers, distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. For example, the picture below shows a 3-degree magic square using the integers of 1 to 9.

Given a finished number square, we need you to judge whether it is a magic square…….

Input

The input contains multiple test cases.

The first line of each case stands an only integer N (0 < N < 10), indicating the degree of the number square and then N lines follows, with N positive integers in each line to describe the number square. All the numbers in the input do not exceed 1000.

A case with N = 0 denotes the end of input, which should not be processed.

Output

For each test case, print “Yes” if it’s a magic square in a single line, otherwise print “No”……………

Sample Input

2
1 2
3 4
2
4 4
4 4
3
8 1 6
3 5 7
4 9 2
4
16 9 6 3
5 4 15 10
11 14 1 8
2 7 12 13
0

Sample Output

No
No
Yes
Yes
先解释一下大体意思，就是先给你一个数字n，然后是n*n的二维数组，然后让你比较任意列、行、对角线的和是否相等。
从题意中我们就知道代码要分4部分了，1.行的和；2.列的和；3.主对角线的和；4.辅对角线的和。
特别注意
特殊处理n=1的情况；
这是我的代码

#include<stdio.h>
#include<string.h>
int main()
{
    int n, i, j, flag, sum1[1000], sum2[1000], a[10][10], sum3, sum4, x;
    while(~scanf("%d", &n))
    {
        memset(sum1, 0, sizeof(sum1));
        memset(sum2, 0, sizeof(sum2));
        flag=0;
        sum3=0;
        sum4=0;
        if(n==0)
        {
            break;
        }
        else if(n==1)
        {
            scanf("%d", &x);
            printf("Yes\n");
            continue;
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                scanf("%d", &a[i][j]);
            }
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                if(a[i][j]==a[0][0])
                {
                    flag=3;
                    break;
                }
                else
                {
                    flag=0;
                }
            }
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                sum1[i]+=a[i][j];
            }
        }
        for(i=0;i<n;i++)
        {
            if(sum1[i]!=sum1[0])
            {
                flag=3;
                break;
            }
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                sum2[i]+=a[j][i];
            }
        }
        for(i=0;i<n;i++)
        {
            if(sum2[i]!=sum2[0])
            {
                flag=3;
                break;
            }
        }
        for(i=0; i<n; i++)
        {
            sum3+=a[i][i];
        }
        for(i=0; i<n; i++)
        {
            sum4+=a[i][n-i-1];
        }
        if(flag==3)
        {
            printf("No\n");
        }
        else
        {
            for(i=0; i<n; i++)
            {
                if(sum1[i]!=sum2[i])
                {
                    flag=4;
                    break;
                }
                else
                {
                    flag=0;
                }
            }
            if((flag==0)&&(sum3==sum4))
            {
                printf("Yes\n");
            }
            else
            {
                printf("No\n");
            }
        }
    }
    return 0;
}

代码看着有点长，但是思路知道了，作对就so easy了。