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。