HDU 1013 Digital Roots 大数水题

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本文介绍了一种计算数字根的方法，通过不断累加整数的各位数字直至得到一位数的过程。提供了完整的AC代码实现，并讨论了一些涉及的数学原理。

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Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73352 Accepted Submission(s): 22899

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

Sample Output

6
3

Source

Greater New York 2000

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一开始以为是水题，虽然没给具体几位数，试着去水了一下，直接WA了，诶，然后就换了大数，那个sum的地方也是试着去试一下，没想到是正确的理论，嗯，一会去研究下到底怎么回事，这个题大数就可以过，其他的都有问题

ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int main()
{
	char s[1005];
	int i,sum;
	
	while(gets(s))
	{
		//puts(s);
		if(s[0]=='0')
			break;
		sum=0;
		for(i=0;s[i]!='\0';i++)
		{
			sum+=(s[i]-'0');
			if(sum>9)//这里设计到数论中的一个知识，叫啥啥啥不变，嗯，记住就好，嘿嘿 
			{
				int n=sum;
				sum=0;
				while(n)
				{
					sum+=n%10;
					n=n/10;
				}
			}
		}
		printf("%d\n",sum);
		
	}
	
	return 0;
}