Add Digits - 俊彥博客

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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 <iostream>

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<<n;
    return 0;
}

#include <iostream>

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<<n;

return 0;

}

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 <iostream>

using namespace std;

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

#include <iostream>

using namespace std;

int main()

{

int n,a;

cin>>n;

if (n==0)

a=0;

else

a=(n-1)%9+1;

cout<<a;

return 0;

}