HDU2135 Rolling table【水题】

Rolling table

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2024    Accepted Submission(s): 977

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

题目大意：给你一个N*N的字符矩阵，再给一个翻转次数M，翻转一次为90°，M为正

表示顺时针翻转，M为负表示逆时针旋转。

思路：将M对4取余，总共分四种情况，根据翻转情况输出相应结果。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

char s[12][12];
int main()
{
    int M,N;
    while(cin >> N >> M)
    {
        getchar();
        memset(s,0,sizeof(s));
        for(int i = 0; i < N; ++i)
            cin >> s[i];
        if(M%4==1 || M%4==-3)
        {
            for(int j = 0; j < N; ++j)
            {
                for(int i = N-1; i >= 0; --i)
                    cout << s[i][j];
                cout << endl;
            }

        }
        else if(M%4 == 2 || M%4==-2)
        {
            for(int i = N-1; i >= 0; --i)
            {
                for(int j = N-1; j >= 0; --j)
                    cout << s[i][j];
                cout << endl;
            }
        }
        else if(M%4 == 3 || M%4==-1)
        {
            for(int j = N-1; j >= 0; --j)
            {
                for(int i = 0; i < N; ++i)
                    cout << s[i][j];
                cout << endl;
            }
        }
        else
        {
            for(int i = 0; i < N; ++i)
            {
                for(int j = 0; j < N; ++j)
                    cout << s[i][j];
                cout << endl;
            }
        }
    }

    return 0;
}