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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.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

不能用循環？先寫個用循環的列出幾個結果，再從中搵規律。

#include 

using namespace std;

int main()
{
    int n,a;
    cin>>n;
    while (n>9)
    {
        a=0;
        while (n)
        {
            a+=n%10;
            n/=10;
        }
        n=a;
    }
    cout<
得到examples:
input / output

33 6

34 7

35 8

36 9

37 1

38 2

39 3

40 4

41 5

42 6

43 7

44 8

45 9

46 1
呃，看起來還是很有規律的。。。。於是得出 output = (input - 1) mod 9 + 1 （其實我都唔知點搞出嚟嘅）
於是：
#include 

using namespace std;

int main()
{
    int n,a;
    cin>>n;
    if (n==0)
        a=0;
    else
        a=(n-1)%9+1;
    cout<