zoukankan      html  css  js  c++  java
  • HDU2137:circumgyrate the string

    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
     


     

    水题

    就是把一个字符串转来转去

    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
        char str[100];
        int len,i,n,j;
        while(~scanf("%s%d",str,&n))
        {
            if(n>=8)
            {
                n = n%8;
            }
            else if(n<0)
            {
                n = n%8;
                n = n+8;
                n = n%8;
            }
            len = strlen(str);
            if(n == 0)
                puts(str);
            else if(n == 1)
            {
                for(i = len-1; i>=0; i--)
                {
                    for(j = 0; j<len; j++)
                    {
                        if(j == i)
                        {
                            printf("%c\n",str[j]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
            else if(n == 2)
            {
                for(i = len-1; i>=0; i--)
                {
                    for(j = 0; j<=len/2; j++)
                    {
                        if(j == len/2)
                        {
                            printf("%c\n",str[i]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
            else if(n == 3)
            {
                for(i = 0; i<len; i++)
                {
                    for(j = 0; j<len; j++)
                    {
                        if(j == i)
                        {
                            printf("%c\n",str[len-1-i]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
            else if(n == 4)
            {
                for(i = len-1; i>=0; i--)
                {
                    putchar(str[i]);
                }
                printf("\n");
            }
            else if(n == 5)
            {
                for(i = 0; i<len; i++)
                {
                    for(j = len-1; j>=0; j--)
                    {
                        if(j == i)
                        {
                            printf("%c\n",str[i]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
            else if(n == 6)
            {
                for(i = 0; i<len; i++)
                {
                    for(j = 0; j<=len/2; j++)
                    {
                        if(j == len/2)
                        {
                            printf("%c\n",str[i]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
             else if(n == 7)
            {
                for(i = 0; i<len; i++)
                {
                    for(j = 0; j<len; j++)
                    {
                        if(j == i)
                        {
                            printf("%c\n",str[i]);
                            break;
                        }
                        else
                            printf(" ");
                    }
                }
            }
        }
        return 0;
    }
    


     

  • 相关阅读:
    如何编译Linux内核
    linux启动过程
    linux ifconfig
    Android 4.0 x86安装教程 附带联网参数详细设置
    linux ntfs模块
    Java 入门进阶
    深入理解Java中的String
    Java中字符串string的数据类型
    IDEA设置JVM运行参数
    Java11实战:模块化的 Netty RPC 服务项目
  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3014187.html
Copyright © 2011-2022 走看看