杭电 1013

博客围绕杭电ACM数字根问题展开。介绍了数字根的定义,即正整数各位数字求和,若结果非一位数则继续求和直至为一位数。给出了问题的时间、内存限制,输入输出要求及示例,还提及写此类题可加积分。

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                                     Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 100606    Accepted Submission(s): 31260

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

24

39

0

Sample Output

6

3

写写水题,加加积分

#include<iostream>
#include<iomanip>
#include<algorithm> 
#include<cstring>
#include<sstream> 
#include<cmath>
#define NUM 20010
int onOff[NUM];
using namespace std;
int main()
{
	char input[1001];
	int sum,low,inputIndex;
	while(cin>>input){
		if(input[0]=='0')
			break;
		
		while(1){
			sum=0;
			for(int i=0; input[i]!='\0'; i++)
				sum = sum + input[i]-'0';
			if(sum<10)
				break;
			//再次封装input
			inputIndex=0;
			while(sum){
				input[inputIndex]='0'+(sum%10);
				inputIndex++;
				sum/=10;
			} 
			input[inputIndex]='\0';
		}
		cout << sum << endl;
	} 
	return 0;
}

 

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