每日一题(15)——Digit Root（大数）

Digital Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23650		Accepted: 7843

 

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

 

本来以为这道题水的不能再水了，可是看了网上的解答才知道，原来这道题的考点在大数上，有可能输入一个上千位的数字……

吸取教训，要仔细考虑问题的细节。

 

[cpp]  view plain copy print ?

1. #include <iostream>  
2.   
3. using namespace std;  
4.   
5. int root(int k)  
6. {  
7.     if(k<10) return k;  
8.     int tmp=0;  
9.     while(k)  
10.     {  
11.         tmp+=k%10;  
12.         k=k/10;  
13.     }  
14.     if(tmp>=10) return root(tmp);  
15.     else return tmp;  
16. }  
17.   
18. int main()  
19. {  
20.     unsigned int n;  
21.     char input[2000];//这题主要考察大数处理，可能有上千位数字,所以应该把输入当成字符串处理   
22.     while(cin>>input, strcmp(input, "0"))//注意字符操作1.  
23.   
24.     {  
25.         n=0;  
26.         for(int i=0; i<strlen(input); i++) n+= input[i] - '0';//注意字符操作2.  
27.         if(n==0)  
28.             break;  
29.         cout<<root(n)<<endl;  
30.     }  
31. }