HDU 2135 Rolling table(方阵旋转+找规律)

Rolling table

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2404    Accepted Submission(s): 1157

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

题解：就是将数的方阵旋转，每次90度。。。。开数组不能用int类型......坑啊

AC代码：

#include<iostream>  
#include<memory.h>  
#include<cstdlib>  
#include<cstdio>  
#include<cmath>  
#include<cstring>  
#include<string>  
#include<cstdlib>  
#include<iomanip>  
#include<vector>  
#include<list>  
#include<map>  
#include<algorithm>  
typedef long long LL;  
using namespace std; 
const LL maxn=1000+10;
const LL mod=10000000; 
char a[12][12]; //坑啊。。。要用char,不能用int 
int main()
{
  int n,m;
  while(cin>>n>>m)
  {   
      
    for(int i=1;i<=n;i++){
       for(int j=1;j<=n;j++)
     {
      cin>>a[i][j];
     }
    }
     int k=m%4;
     if(k==0){
      for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
          cout<<a[i][j];
        }
        cout<<endl;
      }
     }
     if(k==1||k==-3){
      for(int j=1;j<=n;j++)
      {
        for(int i=n;i>=1;--i)
        {
          cout<<a[i][j];
        }
        cout<<endl;
      }
     }
     if(k==2||k==-2)
        {
          for(int i=n;i>=1;i--)
          {
            for(int j=n;j>=1;j--)
            {
              cout<<a[i][j];
            }
            cout<<endl;
          }
        }
        if(k==3||k==-1)
        {
          for(int j=n;j>=1;j--)
          {
            for(int i=1;i<=n;i++)
            {
              cout<<a[i][j];
            }
            cout<<endl;
          }
      }
  }
  return 0;
  
}