HDU1013——Digital Roots

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 69915    Accepted Submission(s): 21885

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

24390

 

Sample Output

63

 
解：题意为给定一个整数，如果该数是1位，则直接输出，两位或者两位以上，则把每一位相加，直到相加和为一位。
因为题目中并没有给定数的范围，所以可能无限大，使用字符串输入进行操作。

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int main(){string a;while(cin>>a && a!="0"){int sum=0;for(int i=0;i<a.length();i++){sum+=a[i]-'0';if(sum>9){sum=sum/10+sum%10;}}printf("%d\n",sum);}return 0;}