hdu 2135

Rolling table

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1318    Accepted Submission(s): 650

Problem Description

After the 32nd ACM/ICPC regional contest, Wiskey is beginning to prepare for CET-6. He has an English words table and read it every morning.
One day, Wiskey's chum wants to play a joke on him. He rolling the table, and tell Wiskey how many time he rotated. Rotate 90 degrees clockwise or count-clockwise each time.
The table has n*n grids. Your task is tell Wiskey the final status of the table.

 

Input

Each line will contain two number.
The first is postive integer n (0 < n <= 10).
The seconed is signed 32-bit integer m.
if m is postive, it represent rotate clockwise m times, else it represent rotate count-clockwise -m times.
Following n lines. Every line contain n characters.

 

Output

Output the n*n grids of the final status.

 

Sample Input

  

   3 2
123
456
789
3 -1
123
456
789
  

 

Sample Output

  

   987
654
321
369
258
147
  

 

Author

Wiskey

 

Source

HDU 2007-11 Programming Contest_WarmUp

 

Recommend

威士忌

 

#include <stdio.h>
int main()
{
	char a[11][11];
	int n, r, i, j;
	while (~scanf("%d%d", &n, &r))
	{
		for(i = 0; i < n; i++)
			scanf("%s", a[i]);
		if(r < 0)
		{
			r *= -1;
			r %= 4;
			r = 4 - r;
		}
		r %= 4;
		if(r == 0)
			for(i = 0; i < n; i++)
				printf("%s\n", a[i]);
		else if(r == 1)
			for(i = 0; i < n; i++)
			{
				for(j = n-1; j >= 0; j--)
					printf("%c", a[j][i]);
				printf("\n");
			}
		else if(r == 2)
			for(i = n-1; i >= 0; i--)
			{
				for(j = n-1; j >= 0; j--)
					printf("%c", a[i][j]);
				printf("\n");
			}
		else if(r == 3)
			for (i = n-1; i >= 0; i--) 
			{
				for(j = 0; j < n; j++)
					printf("%c", a[j][i]);
				printf("\n");
			}
	}
	return 0;
}