Digital Roots—HDU1013 2016-05-06 10:25 85人阅读 评论(0) 收藏

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 67689    Accepted Submission(s): 21144

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.

 

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

 

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题目的意思是累加每位和，若结果大于9，则继续累加

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
char a[1000005];
int main()
{
	int k, sum,sum1;
	while (~scanf(" %s", a))
	{
		if (strcmp(a, "0") == 0)
			break;
		k = strlen(a);
		sum = 0;
		for (int i = 0; i < k; i++)
		{
			sum = sum + a[i] - '0';
		}

		while (sum>9)
		{
			sum1 = 0;
			while (sum != 0)
			{
				sum1 = sum1 + sum % 10;
				sum /= 10;
			}
			sum = sum1;
		}
		printf("%d\n", sum);

	}
	return 0;

}