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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
********************************************/ 参考别的程序,思想为:he problem, widely known as digit root problem, has a congruence formula:
For base b (decimal case b = 10), the digit root of an integer is:
dr(n) = 0 if n == 0
dr(n) = (b-1) if n != 0 and n % (b-1) == 0 dr(n) = n mod (b-1) if n % (b-1) != 0 ordr(n) = 1 + (n - 1) % 9
Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).int addDigits(int num) { return 1 + (num - 1) % 9; }
自己的程序,容易理解:
int addDigits(int num) { while (num>9) num=num/10+num%10; return num; }
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