V - Digital Roots-优快云博客

本文链接：https://blog.youkuaiyun.com/ZJWSA/article/details/74912465

本文介绍了一种计算数字根的方法，即通过不断累加整数的各个位直至得到一个一位数的过程。并提供了一个C++实现的例子，用于处理一系列正整数，输出每个整数对应的数字根。

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题目如下：

The digital root of a positive integer is found bysumming the digits of the integer. If the resulting value is a single digit thenthat digit is the digital root. If the resulting value contains two or moredigits, those digits are summed and the process is repeated. This is continuedas long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields avalue of 6. Since 6 is a single digit, 6 is the digital root of 24. Nowconsider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 isnot a single digit, the process must be repeated. Adding the 1 and the 2 yeilds3, a single digit and also the digital root of 39.

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输入如下：

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 ofzero.

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输出如下：

For each integer in the input, output its digital root ona separate line of the output.

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测试用例输入如下：

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测试用例输出如下：

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结束条件搞错了，长度为1的，耽误了半天。注意点就是，sum的值，可能各位加起来还大于10哦，比如888888,---->48----->12------>3的。

代码如下：

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

using namespace std;

int main()
{
	char a[10005];
	while(gets(a)!=NULL)
	{
		if(strlen(a)==1 && a[0]=='0') break;
	    int sum=0;
		for(int i=0;i<strlen(a);i++){
			sum+=a[i]-'0';
		}
		while(sum>=10)
		{
		    int sum1 = sum;
		    sum=0;
		    while(sum1>0)
			{
		       int t = sum1%10;
			   sum+=t;
			   sum1 = sum1/10;	
		    }
		}
		cout<<sum<<endl;
	}
	return 0;
}

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