zoukankan      html  css  js  c++  java
  • Digital Roots:高精度

    C - Digital Roots

    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.

    Input

    The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

    Output

    For each integer in the input, output its digital root on a separate line of the output.

    Sample Input

    24
    39
    0

    Sample Output

    6
    3

    题目描述:多组输入,输入n,把n的每一位加起来,求出新的n来,不断地加,直到n小于10.用到九余定理。

    #include<iostream>
    #include<string>
    using namespace std;
    int main() {
    	string s;
    	while (cin>>s) {
    		int ans = 0;
    		int lens = s.size();
    		for (int i = 0; i < lens; i++) {
    			ans += s[i] - '0';
    		}
    		if (ans == 0)break;
    		else
    			cout << (ans- 1) % 9 + 1 << endl;		//九余定理 
    	}
    	return 0;
    }
    
    /*
    设x是一个四位数 
    x=a*1000+b*100+c*10+d	=>	x=a*999+b*99+c*9+d+a+b+c	=>要求a+b+c+d 即求 (a*999+b*99+c*9+d+a+b+c)%9即可, 即求x%9 
    特例x==9时,,所以先-1,取余9后再+1 
     例如45=4+5=9
     如果9%9==0		但是结果是0		 
    */ 
    
  • 相关阅读:
    Android 传感器应用
    WebStrom9 体验nodejs
    Web前端框架 小记
    接入淘宝API(PHP版本)
    Android SDK 国内镜像
    Ubuntu14.04 搭建 node.js 环境(Binaries方式)
    C# 异常类型及对应异常类
    .net中序列化读写xml方法的总结
    ASp.NET Core Centos7运行环境搭建
    Linux Centos 常用命令
  • 原文地址:https://www.cnblogs.com/52dxer/p/10548305.html
Copyright © 2011-2022 走看看