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 · 267 阅读

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本文介绍了一个字符串旋转的问题，通过输入一个字符串和一个整数N，将该字符串按逆时针方向旋转N次，每次45度，并展示旋转后的效果。文章提供了一段C++代码实现，该算法能够处理长度不超过80且始终为奇数的字符串。

circumgyrate the string

Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4534 Accepted Submission(s): 1053

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

Author

wangye

Source

HDU 2007-11 Programming Contest_WarmUp

模拟控制格式输出即可

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

int main()
{
    char s[85]={};
    int n,k;
    while(scanf("%s%d",s,&n)!=EOF)
    {
    int len=strlen(s);
    k=n%8;
    if(k<0)
        k+=8;
    if(k==2)
    {
        for(int i=len-1;i>=0;i--)
        {
            printf("%*c\n",(len+1)/2,s[i]);
        }
    }
    else if(k==4)
    {
        for(int i=len-1;i>=0;i--)
        {
            printf("%c",s[i]);
        }
        printf("\n");
    }
    else if(k==6)
    {
        for(int i=0;i<len;i++)
        {
            printf("%*c\n",(len+1)/2,s[i]);
        }
    }
    else if(k==0)
    {
        printf("%s\n",s);
    }
    else if(k==1)
    {
        for(int i=len-1,j=len;i>=0;i--,j--)
        {
            printf("%*c\n",j,s[i]);
        }
    }
    else if(k==3)
    {
        for(int i=len-1,j=1;i>=0;i--,j++)
        {
            printf("%*c\n",j,s[i]);
        }
    }
    else if(k==5)
    {
        for(int i=0;i<len;i++)
        {
            printf("%*c\n",len-i,s[i]);
        }
    }
    else if(k==7)
    {
        for(int i=0;i<len;i++)
        {
            printf("%*c\n",i+1,s[i]);
        }
    }
    }
    return 0;
}