HDU 2137 circumgyrate the string

最新推荐文章于 2022-12-07 14:44:37 发布

原创最新推荐文章于 2022-12-07 14:44:37 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

HDU 同时被 2 个专栏收录

43 篇文章

订阅专栏

字符串处理

0 篇文章

订阅专栏

本文介绍了一种字符串旋转的算法实现，通过模拟字符串绕其中心旋转45度的方式进行N次旋转，适用于长度为奇数的字符串，并提供了完整的C语言代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

把题意搞懂，就是一个字符串的旋转，然后模拟一遍就好~

circumgyrate the string

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char name[100];
int n;

int main()
{
    while( scanf("%s",name) == 1 ){
           int t = strlen(name);
           scanf("%d",&n);
           n %= 8;
           if( n < 0 )
               n += 8;
           if( n == 0 )
               printf("%s\n",name);
           else if( n==4 ){
                for( int i=t-1 ; i>=0 ; i-- )
                     printf("%c",name[i]);
                printf("\n");     
           }
           else if( n == 1 ){
                for( int i=t-1; i>=0 ; i-- ){
                     for( int j=i ; j>0 ; j-- )
                          printf(" ");
                     printf("%c\n",name[i]);
                }
           }
           else if( n == 2 ){
                for( int i=t-1 ; i>=0 ; i-- ){
                     for( int j=0 ; j<(t/2) ; j++ )
                          printf(" ");
                printf("%c\n",name[i]);     
                }
           }
           else if( n == 3 ){
                for( int i=t-1 ; i>=0 ; i-- ){
                     for( int j=t-1 ; j>i ; j-- )
                          printf(" ");
                     printf("%c",name[i]);
                     printf("\n");
                }
           }
           else if( n == 5 ){
                for( int i=0 ; i<t ; i++ ){
                     for( int j=i+1 ; j<t ; j++ )
                          printf(" ");
                     printf("%c\n",name[i]);
                }
           }
           else if( n == 6 ){
                for( int i=0 ; i<t ; i++ ){
                     for( int j=0 ; j<t/2 ; j++ )
                          printf(" ");
                printf("%c\n",name[i]);    
                } 
           }
           else if( n == 7 ){
                for( int i=0 ; i<t ; i++ ){
                     for( int j=0 ; j<i ; j++ )
                          printf(" ");
                     printf("%c\n",name[i]);                           
                }     
           }
    }    
    return 0;
}