HDOJ(HDU) 2135 Rolling table

谙忆

发布于 2021-01-21 15:40:47

1930

发布于 2021-01-21 15:40:47

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

(注意是字符啊！字符！我开始以为是n*n个1-n*n的数字。。。WA了几次) 输入n*n个字符，每次旋转以90°为单位，m>0表示顺时针旋转 m<0逆时针旋转。矩阵旋转后存在四种状态。所以将m取余4，然后判断是哪种状态，然后旋转即可。

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n = sc.nextInt();
            int m = sc.nextInt();
            String strs[]=new String[n];
            for(int i=0;i<n;i++){
                strs[i]=sc.next();
            }
            if(m%4==0){
                for(int i=0;i<n;i++){
                    System.out.println(strs[i]);
                }
                continue;
            }
            if(m>0){
                m=m%4;
                if(m==1){
                    for(int i=0;i<n;i++){
                        for(int j=n-1;j>=0;j--){
                            System.out.print(strs[j].charAt(i));
                        }
                        System.out.println();
                    }
                    continue;
                }
                if(m==2){
                    for(int i=n-1;i>=0;i--){
                        for(int j=n-1;j>=0;j--){
                            System.out.print(strs[i].charAt(j));
                        }
                        System.out.println();
                    }
                    continue;
                }
                if(m==3){
                    for(int i=n-1;i>=0;i--){
                        for(int j=0;j<n;j++){
                            System.out.print(strs[j].charAt(i));
                        }
                        System.out.println();
                    }
                    continue;
                }
            }
            if(m<0){
                m=m*(-1);
                m=m%4;
                if(m==3){
                    for(int i=0;i<n;i++){
                        for(int j=n-1;j>=0;j--){
                            System.out.print(strs[j].charAt(i));
                        }
                        System.out.println();
                    }
                    continue;
                }
                if(m==2){
                    for(int i=n-1;i>=0;i--){
                        for(int j=n-1;j>=0;j--){
                            System.out.print(strs[i].charAt(j));
                        }
                        System.out.println();
                    }
                    continue;
                }
                if(m==1){
                    for(int i=n-1;i>=0;i--){
                        for(int j=0;j<n;j++){
                            System.out.print(strs[j].charAt(i));
                        }
                        System.out.println();
                    }
                }
            }
        }
    }
}

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原始发表：2016/05/03 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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