14周 — Digital Roots

最新推荐文章于 2021-07-14 19:55:43 发布

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数字根是一个正整数通过不断将各位数字相加直至得到一位数的过程。例如，24的数字根是6（2+4），39的数字根是3（3+9→12→1+2）。本文介绍如何处理超过2的64次方的输入，使用数组而非长整型来解决这个问题。示例输入包括24、39，输出分别为它们的数字根6和3。

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

实际上就是把数拆开加一直加到个位，最开始还是个瓜皮，以为long long int就能表示2的64次方，结果不行，，拿数组就好了，，，，，，还有就是strcmp里的0是双引号，

#include <stdio.h>
#include <string.h>
int f(int x)
{
	int sum=0;
	while(x!=0)
	{
		sum+=x%10;
		x/=10;
	}
	return sum;
}
int main()
{
	char a[100000];
	while(scanf("%s",a) !=EOF)
	{
		if(strcmp(a,"0")==0) break;    
		int cd=strlen(a);
		int jg=0,gc=0;
		for(int i=0;i<cd;i++)
			jg+=a[i]-48;						//把最开始的写成一个整数
		gc=f(jg);
		while(gc>=10)
		{
			gc=f(gc);
		}
		printf("%d\n",gc);
	}
	return 0;
}