2015
09-17

In the ancient three kingdom period, Zhuge Liang was the most famous and smart military leader. His enemy was Sima Yi, the military leader of Kingdom Wei. Sima Yi always looked stupid when fighting against Zhuge Liang. But it was Sima Yi who laughed to the end.
Once, Sima Yi defeated Zhuge Liang again, and got many WBFHs in the battle field. But he didn’t know the passwords. Ma Su’s son betrayed Zhuge Liang and came to Sima Yi. He told Sima Yi the way to figure out the password by two cards.He said to Sima Yi:
"A password card is a square grid consisting of N×N cells.In each cell,there is a number. Two password cards are of the same size. If you overlap them, you get two numbers in each cell. Those two numbers in a cell may be the same or not the same. You can turn a card by 0 degree, 90 degrees, 180 degrees, or 270 degrees, and then overlap it on another. But flipping is not allowed. The maximum amount of cells which contains two equal numbers after overlapping, is the password. Please note that the two cards must be totally overlapped. You can’t only overlap a part of them."
Now you should find a way to figure out the password for each WBFH as quickly as possible.

There are several test cases.
In each test case:
The first line contains a integer N, meaning that the password card is a N×N grid(0<N<=30).
Then a N×N matrix follows ,describing a password card. Each element is an integer in a cell.
Then another N×N matrix follows, describing another password card.
Those integers are all no less than 0 and less than 300.
The input ends with N = 0

There are several test cases.
In each test case:
The first line contains a integer N, meaning that the password card is a N×N grid(0<N<=30).
Then a N×N matrix follows ,describing a password card. Each element is an integer in a cell.
Then another N×N matrix follows, describing another password card.
Those integers are all no less than 0 and less than 300.
The input ends with N = 0

2
1 2
3 4
5 6
7 8
2
10 20
30 13
90 10
13 21
0

0
2

AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
int a[31][31],b[31][31];
int main()
{
int N;int i,j,m,n;
while(scanf("%d",&N)!=EOF,N)
{
int cnt[5]={0};
//cout<<1<<endl;
for( i=1;i<=N;i++)
for( j=1;j<=N;j++)
scanf("%d",&a[i][j]);
for( i=1;i<=N;i++)
for( j=1;j<=N;j++)
scanf("%d",&b[i][j]);
for(i=1;i<=N;i++)
{
for(j=1;j<=N;j++)
{
if(a[i][j]==b[i][j])
cnt[1]++;
if(a[N+1-j][i]==b[i][j])
cnt[2]++;
if(a[N+1-i][N+1-j]==b[i][j])
cnt[3]++;
if(a[j][N+1-i]==b[i][j])
cnt[4]++;
}
//cout<<-1<<endl;
}
int sum=0;
for(i=1;i<=4;i++)
{
sum=max(sum,cnt[i]);
}
printf("%d\n",sum);
}
}