ZOJ 1115 Digital Roots （C）

原创于 2019-08-31 10:06:30 发布 · 178 阅读

CC 4.0 BY-SA版权

本文介绍了一种计算数字根的算法，数字根是指将一个正整数的各位数字反复相加直至得到一位数的过程。文章提供了详细的算法思路和C语言实现代码，通过将数字视为字符数组输入，对数组中的每个数字进行求和，若和大于等于10则继续求和，直至得到一位数。

题目
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.注意！！该题的输入可能超过2的64次方！！

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

题意：把一个数的各个位数上的数相加，直到小于10
代码思路：把数字看做数组输入，把数组中的各个数字相加，如果sum满足要求，则输出，否则对9取余

#include <stdio.h>
#include <string.h>
int main()
{
	char n[1000010];
	long long sum=0,i,len;
	gets(n);
	while(n[0]!='0') 	
	{
		len=strlen(n);
		for(i=0;i<len;i++)
		    sum=sum+(n[i]-'0'); 
		if(sum>0&&sum<10)
		    printf("%lld\n",sum);
		else
		{
			sum=sum%9;  
    		        if(sum==0) printf("9\n");
	    	        else printf("%lld\n",sum);
		}
		sum=0;
		gets(n);
	}
	return 0;
}