zoukankan      html  css  js  c++  java
  • 【hdu 1163 Eddy's digital Roots 】

    Eddy's digital Roots

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2355    Accepted Submission(s): 1348


    Problem Description
    The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

    For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

    The Eddy's easy problem is that : give you the n,want you to find the n^n's digital Roots.
     
    Input
    The input file will contain a list of positive integers n, one per line. The end of the input will be indicated by an integer value of zero. Notice:For each integer in the input n(n<10000).
     
    Output
    Output n^n's digital root on a separate line of the output.
     
    Sample Input
    2 4 0
     
    Sample Output
    4 4
     
    Author
    eddy
     
    Recommend
    JGShining
     
     
     
     1 // Project name : 1163 ( Eddy's digital Roots ) 
     2 // File name    : main.cpp
     3 // Author       : Izumu
     4 // Date & Time  : Sun Jul  8 20:59:43 2012
     5 
     6 
     7 #include <iostream>
     8 using namespace std;
     9 
    10 int main()
    11 {
    12     int n;
    13     while (cin >> n && n)
    14     {
    15         int sum = 1;
    16         for (int i = 0; i < n; i++)
    17         {
    18             sum = sum * n % 9;
    19         }
    20 
    21         if (sum == 0)
    22         {
    23             cout << "9" << endl;
    24         }
    25         else
    26         {
    27             cout << sum << endl;
    28         }
    29     }
    30     return 0;
    31 }
    32 
    33 // end 
    34 // ism 
  • 相关阅读:
    hadoop之 解析HDFS的写文件流程
    Linux之 手动释放内存
    Heka 的编译
    go get 下载需要的相关工具
    峰值计算的方法
    thrift简介
    Bazaar 版本控制工具
    Homebrew
    虚拟机下centos时间不正确的方便解决方法
    golang 应用的部署相关技术
  • 原文地址:https://www.cnblogs.com/ismdeep/p/2581857.html
Copyright © 2011-2022 走看看