Magic Square
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
A magic square is a 3×3 square, where each element is a single digit between 1 and 9 inclusive, and each digit appears exactly once. There are 4 different contiguous 2×2subsquares in a magic squares, which are labeled from 1 to 4 as the following figure shows. These 2×2 subsquares can be rotated. We use the label of the subsquare with an uppercase letter to represent a rotation. If we rotate the subsquare clockwise, the letter is 'C'; if we rotate it counterclockwise, the letter is 'R'. The following figure shows two different rotations.
Now, given the initial state of a magic square and a sequence of rotations, please print the final state of the magic square after these rotations are performed.
Input
The first line of input is a single integer T (1≤T≤100), the number of test cases.
Each test case begins with a single integer n (1≤n≤100), the number of rotations. It is then followed by a 3×3 square, where every digit between 1 and 9 inclusive appears exactly once, representing the initial state of the magic square. The following n lines describe the sequence of rotations.
The test data guarantees that the input is valid.
Output
For each test case, display a 3×3 square, denoting the final state of the magic square.
Sample Input
1
2
123
456
789
1C
4R
Sample Output
413
569
728
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
using namespace std;
char matrx[4][4];
int x,y;
void work()
{
swap(matrx[x][y],matrx[x][y+1]);
swap(matrx[x+1][y],matrx[x+1][y+1]);
swap(matrx[x][y],matrx[x+1][y+1]);
}
void work1()
{
swap(matrx[x][y],matrx[x+1][y]);
swap(matrx[x][y+1],matrx[x+1][y+1]);
swap(matrx[x][y],matrx[x+1][y+1]);
}
int main()
{
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
getchar();
for(int i=1; i<=3; i++)
{
for(int j=1; j<=3; j++)
scanf("%c",&matrx[i][j]);
getchar();
}
char ch;
int t;
for(int i=0; i<n; i++)
{
scanf("%d%c",&t,&ch);
if(t==1)
{
x=1;
y=1;
}
else if(t==2)
{
x=1;
y=2;
}
else if(t==3)
{
x=2;
y=1;
}
else
{
x=2;
y=2;
}
if(ch=='C')
work();
else
work1();
}
for(int i=1; i<=3; i++)
{
for(int j=1; j<=3; j++)
printf("%c",matrx[i][j]);
printf("\n");
}
}
return 0;
}