杭电 1013 Digital Roots

这题嘛，不用多说，难度不大，唯一要注意的，也是ACM经常耍的花招：1000以内的单次输入量。也就是说，它可能输入1000 个数字字符，直观点举例，就是：

8947129387523745194375983274597435792834527345297349594375972345414132413577230

诸如此类。

所以必须用字符串来解决。

/* THE PROGRAM IS MADE BY PYY */ /*---------------------------------------------------------- http://acm.hdu.edu.cn/showproblem.php?pid=1013 Digital Roots Begin : 15:30 End : 15:57 ----------------------------------------------------------*/ #include <iostream> using namespace std; int main() { int sum, tmp, i; char c; while (cin >> c && c != '0') { sum = c - '0'; while (cin.get(c) && c != '/n') { sum += c - '0'; } tmp = sum; sum = 0; while (tmp) { sum += tmp % 10; tmp /= 10; if (!tmp && sum > 9) { tmp = sum; sum = 0; } } cout << sum << endl; } return 0; }

------------------------------------------原题如下 ---------------------------------------------

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19048Accepted Submission(s): 5487

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

Source

Greater New York 2000