circumgyrate the string

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3028    Accepted Submission(s): 696

Problem Description

  Give you a string, just circumgyrate. The number N means you just   circumgyrate the string N times, and each time you circumgyrate the string for 45 degree anticlockwise.

 

Input

  In each case there is string and a integer N. And the length of the string is always odd, so the center of the string will not be changed, and the string is always horizontal at the beginning. The length of the string will not exceed 80, so we can see the complete result on the screen.

 

Output

  For each case, print the circumgrated string.

 

Sample Input

asdfass 7

 

Sample Output

a s  d   f    a     s      s

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest_WarmUp

 

<转>

#include<stdio.h>#include<string.h>int main(){    char a[88];    int n,i,j,k;    while(scanf("%s",a)!=EOF)    {        k=strlen(a);        scanf("%d",&n);        n=n%8;        if(n<0)            n=n+8;        switch(n)        {        case 0:        {             for(i=0;i<k;i++)                 printf("%c",a[i]);             printf("\n");             break;        }        case 1:        {             for(j=k-1;j>=0;j--)             {                 for(i=0;i<j;i++)                     printf(" ");                 printf("%c\n",a[j]);             } break;        }        case 2:        {             for(j=k-1;j>=0;j--)             {                 for(i=0;i<k/2;i++)                     printf(" ");                 printf("%c\n",a[j]);             } break;        }             case 3:        {              for(j=k-1;j>=0;j--)              {                  for(i=k-1;i>j;i--)                      printf(" ");                  printf("%c\n",a[j]);              } break;        }        case 4:        {               for(i=k-1;i>=0;i--)                   printf("%c",a[i]);               printf("\n");               break;        }        case 5:        {             for(j=0;j<k;j++)             {                 for(i=k-1;i>j;i--)                     printf(" ");                 printf("%c\n",a[j]);             } break;        }        case 6:        {             for(j=0;j<k;j++)             {                 for(i=0;i<k/2;i++)                     printf(" ");                 printf("%c\n",a[j]);             } break;        }        case 7:        {             for(j=0;j<k;j++)             {                 for(i=0;i<j;i++)                     printf(" ");                 printf("%c\n",a[j]);             } break;        }        default :break;        }    }     return 0;} 

/*我自己写的，总是错，不知道哪里错了，~~~~(>_<)~~~~ */#include<iostream>#include<iomanip>using namespace std;int main(){char a[88];int n;while(cin>>a){cin>>n;int m=n%8,i,j;if(m<0)m=m+8;int s=strlen(a);switch(m){case 0:{for(i=0;i<s;i++)cout<<a[i];cout<<endl;break;   }case 1:{for(i=s-1;i>=0;i--){for(j=0;j<s;j++){if(i==j)cout<<a[i];elsecout<<" ";}cout<<endl;}break;   }case 2:{for(i=s-1;i>=0;i--){for(j=0;j<s/2;j++)cout<<" ";cout<<a[i]<<endl;}break;}case 3:{for(i=s-1;i>=0;i--){for(j=s-1;j>=0;j--){if(i==j)cout<<a[i];elsecout<<" ";}cout<<endl;}break;   }case 4:{for(i=s-1;i>=0;i--)cout<<a[i];cout<<endl;break;}case 5:{for(i=0;i<s;i++){for(j=s-1;j>=0;j--){if(i==j)cout<<a[i];elsecout<<" ";}cout<<endl;}break;   }case 6:{for(i=0;i<s;i++){for(j=0;j<s/2;j++)cout<<" ";cout<<a[i]<<endl;}break;   }case 7:{for(i=0;i<s;i++){for(j=0;j<s;j++){if(i==j)cout<<a[i];elsecout<<" ";}cout<<endl;}break;} default :break;}}return 0;}