题意：
Let’s call some square matrix with integer values in its cells palindromic if it doesn’t change after the order of rows is reversed and it doesn’t change after the order of columns is reversed.
For example, the following matrices are palindromic:
The following matrices are not palindromic because they change after the order of rows is reversed:
The following matrices are not palindromic because they change after the order of columns is reversed:
You are given n2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print “NO”.
Input
The first line contains one integer n (1≤n≤20).
The second line contains n2 integers a1,a2,…,an2 (1≤ai≤1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print “YES”. Then print n lines with n space-separated numbers — the resulting matrix.
If it’s impossible to construct any matrix, then print “NO”.
You can print each letter in any case (upper or lower). For example, “YeS”, “no” and “yES” are all acceptable.
Examples
inputCopy
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
outputCopy
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
inputCopy
3
1 1 1 1 1 3 3 3 3
outputCopy
YES
1 3 1
3 1 3
1 3 1
inputCopy
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
outputCopy
NO
inputCopy
1
10
outputCopy
YES
10
Note
题解：
仔细观察可以看出来，如果n是偶数，那么矩阵没有轴，所以每种数字都必须4个4个出现，如果某个数的个数%4不是0则NO，模拟输出的时候就模拟1/4块就可以了
如果n是奇数，矩阵有轴，那么先模拟非轴的区域，找出4个4个出现的数字去填充；然后模拟轴上的区域，找出若干个剩余数量还在2个以上的数来填充轴就可以了，如果过程中找不出数了就NO
代码：
#include<stdio.h>
int ans[1005][1005];
int main()
{
int n,a[1005],sum[1005],b,k;
scanf("%d",&n);
for(int i=0;i<1005;i++)sum[i]=0;
for(int i=0;i<n*n;i++)
{
scanf("%d",&b);
int flag=0;
for(int j=0;j<k;j++)
{
if(b==a[j]&&sum[j]<4)
{
sum[j]++;
flag=1;
break;
}
}
if(flag==0)
{
a[k]=b;
sum[k]=1;
k++;
}
}
if(n%2==0)
{
for(int i=0;i<k;i++)
{
if(sum[i]!=4)
{
printf("NO\n");
return 0;
}
}
printf("YES\n");
int s=0;
for(int i=0;i<n/2;i++)
{
for(int j=0;j<n/2;j++)
{
ans[i][j]=a[s];
ans[i][n-1-j]=a[s];
ans[n-1-i][j]=a[s];
ans[n-1-i][n-1-j]=a[s];
s++;
}
}
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(j==n-1)printf("%d\n",ans[i][j]);
else printf("%d ",ans[i][j]);
}
}
}
else
{
int s=0;
for(int i=0;i<n/2;i++)
{
for(int j=0;j<n/2;j++)
{
while(sum[s]!=4)
{
s++;
if(s>=k)
{
printf("NO\n");
return 0;
}
}
ans[i][j]=a[s];
ans[i][n-1-j]=a[s];
ans[n-1-i][j]=a[s];
ans[n-1-i][n-1-j]=a[s];
sum[s]=0;
}
}
s=0;
for(int i=0;i<n/2;i++)
{
while(sum[s]<2)
{
s++;
if(s>=k)
{
printf("NO\n");
return 0;
}
}
ans[n/2][i]=a[s];
ans[n/2][n-1-i]=a[s];
sum[s]-=2;
}
s=0;
for(int i=0;i<n/2;i++)
{
while(sum[s]<2)
{
s++;
if(s>=k)
{
printf("NO\n");
return 0;
}
}
ans[i][n/2]=a[s];
ans[n-1-i][n/2]=a[s];
sum[s]-=2;
}
for(int i=0;i<k;i++)if(sum[i])ans[n/2][n/2]=a[i];
printf("YES\n");
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(j==n-1)printf("%d\n",ans[i][j]);
else printf("%d ",ans[i][j]);
}
}
}
}