HDU 2135 Rolling table

http://acm.hdu.edu.cn/showproblem.php?pid=2135

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

 

代码：

#include <bits/stdc++.h>
using namespace std;

int N, M;
char mp[15][15];

int main() {
    while(~scanf("%d%d", &N, &M)) {
        int flag;
        if(M > 0) flag = 1;
        else flag = 0;
        memset(mp, 0, sizeof(mp));
        for(int i = 1; i <= N; i ++) {
            getchar();
            for(int j = 1; j <= N; j ++)
                scanf("%c", &mp[i][j]);
        }

        if(abs(M) % 4 == 0) {
            for(int i = 1; i <= N; i ++) {
                for(int j = 1; j <= N; j ++)
                    printf("%c", mp[i][j]);
                printf("\n");
            }
        } else {
            if(abs(M) % 4 == 2) {
                for(int i = N; i >= 1; i --) {
                    for(int j = N; j >= 1; j --)
                        printf("%c", mp[i][j]);
                    printf("\n");
                }
            } else if(M > 0 && M % 4 == 1 || M < 0 && abs(M) % 4 == 3) {
                for(int j = 1; j <= N; j ++) {
                    for(int i = N; i >= 1; i --)
                        printf("%c", mp[i][j]);
                    printf("\n");
                }
            } else if(M > 0 && M % 4 == 3 || M < 0 && abs(M) % 4 == 1) {
                for(int j = N; j >= 1; j --) {
                    for(int i = 1; i <= N; i ++)
                        printf("%c", mp[i][j]);
                    printf("\n");
                }
            }
        }
    }
    return 0;
}