# Magic Square

Magic Square

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.

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.

For each test case, print "Yes" if it's a magic square in a single line, otherwise print "No".

``````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
``````

``````No
No
Yes
Yes
``````

``````#include<stdio.h>
int cs(int a[],int n)
{
int i,j;
for(i=1;i<=n-1;i++)
{
for(j=i+1;j<=n;j++)
if(a[i]==a[j])return 1;
}
return 0;
}
void pd(int sum[],int n)
{
int s=sum[1],flag=0,i;
for(i=1;i<=2*n+2;i++)
{
if(sum[i]!=s)flag=1;
}
if(flag==0) printf("Yes\n");
else printf("No\n");
}
int main()
{
int n,a[11][11],i,j,sum[50],s[121],k;
while(scanf("%d",&n)&& n!=0)
{
if(n==1)
{
scanf("%d",&a[1][1]);
printf("Yes\n");
continue;
}
k=1;
for(i=1;i<=50;i++)
sum[i]=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
s[k++]=a[i][j];
sum[i]+=a[i][j];
if(i==j)
sum[2*n+1]+=a[i][j];
if(i==n+1-j)
sum[2*n+2]+=a[i][j];
}
}
if(cs(s,n*n))
{
printf("No\n");
continue;
}
for(j=1;j<=n;j++)
for(i=1;i<=n;i++)
sum[i+n]+=a[i][j];
pd(sum,n);
}
return 0;
}
``````

``````#include<stdio.h>
#include<math.h>
int main()
{
int n;
int a[100][100],b[1005],c[50];
int i,j,k;
while(1)
{
scanf("%d",&n);
if(n==0)
break;
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
b[a[i][j]]++;
}
for(i=0;i<1000;i++)
if(b[i]>1)
break;
if(i<1000)
printf("No\n");
else
{
int d1=0,d2=0;
for(i=0;i<n;i++)
{
d1+=a[i][i];
d2+=a[i][n-i-1];
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
c[i]+=a[i][j];
k=i;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
c[k]+=a[j][i];
k++;
}
if(d1!=d2)
printf("No\n");
else
{
for(i=0;i<k;i++)
if(c[i]!=d1)
break;
if(i==k)
printf("Yes\n");
else
printf("No\n");
}
}
}
return 0;

}
``````