Digital Roots

最新推荐文章于 2020-05-10 23:42:17 发布

Kelvinwuxiao

最新推荐文章于 2020-05-10 23:42:17 发布

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分类专栏： Coding 文章标签： C++ acm

本文链接：https://blog.youkuaiyun.com/kelvinwuxiao/article/details/11558101

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本文详细介绍了如何使用C++优化代码来快速计算正整数的数字根，特别关注了处理大数时的技巧。通过将输入转换为字符串并逐位相加，确保了程序在面对极大数据时的高效运行。文章还强调了避免使用基本整型变量以防止溢出的重要性，并提供了一个简洁高效的实现示例。

Background
The digital root of a positive integer is found by summing the digits of theinteger. If the resulting value is a single digit then that digit is thedigital root. If the resulting value contains two or more digits, those digitsare summed and the process is repeated. This is continued as long as necessaryto obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields avalue of 6. Since 6 is a single digit, 6 is the digital root of 24. Nowconsider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 isnot a single digit, the process must be repeated. Adding the 1 and the 2 yeilds3, 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 endof 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 ofthe output.

Example

Input

Output

To be honest, it's really easy one. But when I have finished my coding work, time exceeded !!!! WTF

Time limit is 2s, my running time is 2001ms, just 1ms.

Then I do a lot work to figure it out.

I found the input number maybe extremely BIG. It will reload the function a lot times. So we should not use the int, but the string as input.

the loop is funny, it can circulate infintely.

 for(j=0;;j++)

Also, if you input a string like num, the num[0] will be the first letter. num[0]-'0' will make str become int number

#include <iostream>
#include <string>
using namespace std;

int fuck(int y){
    int r=0;
    while(y!=0){
        r+=y%10;
        y=y/10;
    }
    return r;
}

int main()
{
    int j,b,k;
    string num;
    for(j=0;;j++)
    {
        cin>>num;
        if(num[0]=='0' && num.length()==1) break;
        int sum=0;
        for(k=0;k<num.length();k++){
            sum+=num[k]-'0';
        }
        b=fuck(sum);
        while(b>9){
            b=fuck(b);
        }
        cout<<b<<endl;
    }
        return 0;
}