Digital Roots，函數

 Digital Roots

時間限制: 1 Sec  記憶體限制: 128 MB

送出: 91  解決: 29

[送出][狀态][論壇]

題目描述

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.

輸入

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.

輸出

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

樣例輸入

24

39

樣例輸出

6

3

提示

數字可能很大，如99999999999999999999999999999999999999

#include<iostream>

#include<string.h>

using namespace std;

int t;

int su(int sum)

{

t=0;

    while(sum)

    {

      t+=sum%10;

      sum/=10;

    }

if(t>=10)

 return su(t);

    else 

return t;

}

int main()

    int n,sum;

    char x[1000];

    while(cin>>x&&x[0]!=‘0‘)

        sum=0;

        for(int i=0;i<strlen(x);i++)

        {

            sum+=x[i]-‘0‘;

        }

        cout<<su(sum)<<endl;

定義為數組，每位相加，最後為1位數，即為結果