数根计算ZOJ 1115 Digital Roots

提交地址http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1115

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

6
3

数根的问题从经不知一次的遇到，经常遇到一些不同的做法

先说一下我们常规思维，把每位相加

#include<string.h>
#include<stdio.h>

int main()
{
	char a[1001];
	while(scanf("%s",a)!=EOF)
	{
		if(strcmp(a,"0")==0)
		{
			return 0;
		}
		int sum =0;
		for(int i=0;i<strlen(a);i++)
		{
			sum+=a[i]-‘0‘;
		}
		while(sum/10!=0)
		{
			int t = sum;
			sum=0;
			while(t!=0)
			{
				sum+=t%10;
				t/=10;
			}
		}
		printf("%d\n",sum);
	}
}

我感觉模很神奇，经常遇到，明天写篇关于模运算的。

#include<stdio.h>
#include<string.h>
int main()
{
	char a[1001];
	while(scanf("%s",a)==1)
	{
		if(strcmp(a,"0")==0)
		{
			break;
		}
		int sum =0;
		for(int i=0;i<strlen(a);i++)
		{
			sum+=a[i]-‘0‘;
		}
		if(sum%9==0)
			printf("9\n");
		else
			printf("%d\n",sum%9);
	}
}

感谢自己继续坚持！

数根计算ZOJ 1115 Digital Roots,布布扣,bubuko.com

数根计算ZOJ 1115 Digital Roots

原文：http://blog.csdn.net/u010123208/article/details/24199055