ZOJ Problem Set - 1115 Digital Roots

Digital Roots

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Background

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.

Example

Input

24
39
0

Output

6
3

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Source: Greater New York 2000

分析：

数字与字符串之间的转换，用了最简单的转换方式，效率也不错。

重点是第一轮的转换，因为输入的数据可能非常大，但是进过第一轮转换后，数字肯定很小，至少int肯定能够容纳下。所以后面的转换就是设置一个while循环，看看sum是否小于10，是的话就输出，不是就继续循环，直到小于10.

简单的水题。

ac代码：

#include <iostream>
#include<cstdio>
#include<string>
using namespace std;
string s;
int main()
{
    while(cin>>s&&s[0]!='0')
    {
        int sum=0;
        int i,L=s.length();
        for(i=0;i<L;i++)
        {
            if(s[i]=='1') sum+=1;
            else if(s[i]=='2') sum+=2;
            else if(s[i]=='3') sum+=3;
            else if(s[i]=='4') sum+=4;
            else if(s[i]=='5') sum+=5;
            else if(s[i]=='6') sum+=6;
            else if(s[i]=='7') sum+=7;
            else if(s[i]=='8') sum+=8;
            else if(s[i]=='9') sum+=9;
        }
        while(true)
        {
            if(sum<10)
            {
                printf("%d\n",sum);
                break;
            }
            else sum=sum/10+sum%10;
        }
    }
    return 0;
}