http://codeforces.com/contest/1118/problem/C
题意：给n(0-20),和n*n数组，构造n*n回文数组，行和列调换不变
思路：n为偶时，必然每个数时4的倍数；奇时个数为奇数的一个是最中间，中间行和列两两构造，剩下的四个四个构造；
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e4+9;
const int mod=1e9+7;
template <class T>
inline void sc(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
int num[maxn+10];
int a[100][100];
int main()
{
int n;
cin>>n;
for(int i=0;i<n*n;i++)
{
int x;
cin>>x;
num[x]++;
}
if(n%2==0)
{
for(int i=1;i<=maxn;i++)
{
if(num[i]%4)
{
cout<<"NO"<<endl;
return 0;
}
}
int id=1;
for(int i=0;i<n/2;i++)
{
for(int j=0;j<n/2;j++)
{
while(1)
{
if(num[id])
{
a[i][j] = id;
a[n-i-1][j] = id;
a[i][n-j-1] = id;
a[n-i-1][n-j-1] = id;
num[id] -= 4;
break;
}
id++;
if(id>1000)
{
cout<<"NO"<<endl;
return 0;
}
}
}
}
cout<<"YES"<<endl;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
else
{
int ji=0;
for(int i=1;i<=maxn;i++)
{
if(num[i]%2)
{
ji++;
}
}
if(ji!=1)
{
cout<<"NO"<<endl;
return 0;
}
//处理最中间的
for(int i=1;i<=maxn;i++)
{
if(num[i]%2)
{
a[n/2][n/2]=i;
num[i]--;
break;
}
}
//处理中间行，列
//先用2个的
int cnt=0;
for(int i=1;i<=maxn;i++)
{
if(cnt==n-1) break;
if(num[i]%4)
{
if(cnt<n/2)
{
//cout<<n/2<<" "<<cnt<<" "<<i<<endl<<endl;;
a[n/2][cnt]=i;
a[n/2][n-cnt-1]=i;
num[i]-=2;
}
else
{
//cout<<cnt-n/2<<" "<<n/2<<" "<<i<<endl<<endl;
a[(cnt-n/2)][n/2]=i;
a[n-(cnt-n/2)-1][n/2]=i;
num[i]-=2;
}
cnt++;
}
}
//2个不够用4个的
for(int i=1;i<=maxn;i++)
{
if(cnt==n-1) break;
if(num[i])
{
while(num[i])
{
if(cnt==n-1) break;
if(cnt<n/2)
{
//cout<<n/2<<" "<<cnt<<" "<<i<<endl<<endl;;
a[n/2][cnt]=i;
a[n/2][n-cnt-1]=i;
num[i]-=2;
}
else
{
//cout<<cnt-n/2<<" "<<n/2<<" "<<i<<endl<<endl;
a[(cnt-n/2)][n/2]=i;
a[n-(cnt-n/2)-1][n/2]=i;
num[i]-=2;
}
cnt++;
}
}
}
/*for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
cout<<a[i][j]<<" ";
}
cout<<endl;
}*/
int id=1;
for(int i=0;i<n/2;i++)
{
for(int j=0;j<n/2;j++)
{
while(1)
{
if(num[id]%4==0&&num[id])
{
a[i][j] = id;
a[n-i-1][j] = id;
a[i][n-j-1] = id;
a[n-i-1][n-j-1] = id;
num[id] -= 4;
break;
}
id++;
if(id>1000)
{
cout<<"NO"<<endl;
return 0;
}
}
}
}
cout<<"YES"<<endl;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
return 0;
}
python：
纯粹学习，运行时间很长，学会了构造二维数组
maxn=10009
num=[0]*maxn
a=[([0 for i in range(maxn)])for i in range(maxn)]
n=int(input())
mid=int(n/2)
op=input()
op=op.split()
for i in range(len(op)):
op[i]=int(op[i])
num[op[i]]+=1
if n%2==0 :
for i in range(1,maxn):
if num[i]%4:
print("NO")
exit()
id=1
for i in range(mid):
for j in range(mid):
while 1:
if num[id]:
a[i][j]=id
a[n-i-1][j]=id
a[i][n - j - 1] = id
a[n - i - 1][n - j - 1] = id
num[id] -= 4
break
id+=1
if id>1000:
print("NO")
exit()
print('YES')
for i in range(n):
for j in range(n):
print(a[i][j],end=' ')
print()
else:
ji=0
for i in range(1,maxn):
if num[i]%2:
ji+=1
if ji!=1:
print('NO')
exit()
for i in range(1,maxn):
if num[i]%2:
a[mid][mid]=i
num[i]-=1
break
cnt=0
for i in range(1,maxn):
if cnt==n-1:
break
if num[i]%4:
if cnt<mid:
a[mid][cnt]=i
a[mid][n-cnt-1]=i
num[i]-=2
else:
a[cnt-mid][mid]=i
a[n-(cnt-mid)-1][mid]=i
num[i]-=2
cnt+=1
for i in range(1,maxn):
if cnt==n-1:
break
if num[i]:
while num[i]:
if cnt==n-1:break
if cnt<mid:
a[mid][cnt]=i
a[mid][n-cnt-1]=i
num[i]-=2
else:
a[cnt-mid][mid]=i
a[n-(cnt-mid)-1][mid]=i
num[i]-=2
cnt+=1
id=1
for i in range(0,mid):
for j in range(0,mid):
while 1:
if(num[id]%4==0 and num[id]):
a[i][j] = id
a[n - i - 1][j] = id
a[i][n - j - 1] = id
a[n - i - 1][n - j - 1] = id
num[id] -= 4
break
id+=1
if id>1000:
print('NO')
exit()
print('YES')
for i in range(0,n):
for j in range(0,n):
print(a[i][j],end=' ')
print()