# UVA 11332 Summing Digits
## 題目連結 [UVA 11332](https://vjudge.net/problem/UVA-11332)
### 題目內容
For a positive integer n, let f(n) denote the sum of the digits of n when represented in base 10. It is easy to see that the sequence of numbers n, f(n), f(f(n)), f(f(f(n))), . . . eventually becomes a single digit number that repeats forever. Let this single digit be denoted g(n).
For example, consider n = 1234567892.
Then:
f(n) = 1+2+3+4+5+6+7+8+9+2 = 47
f(f(n)) = 4 + 7 = 11
f(f(f(n))) = 1 + 1 = 2
Therefore, g(1234567892) = 2
### 輸入限制
Each line of input contains a single positive integer n at most 2,000,000,000. Input is terminated by n = 0 which should not be processed.
### 輸出限制
For each such integer, you are to output a single line containing g(n).

### 解題思路
將數字當陣列算到長度為1為止。
### 程式碼

#include<bits/stdc++.h>
using namespace std;
int main(){
    string n;
    while(cin>>n && n!="0"){
        long long ans=0;
        while(n.length()!=1){
            ans=0;
            for(int i=0;i<n.length();i++){
                ans+=n[i]-'0';
            }
            n=to_string(ans);
        }
        cout<<n<<endl;
    }
}

## 測資
### Sample input
2
11
47
1234567892
0
### Sample output
2
2
2
2
## 中文題目連結 [zerojudge c813](https://zerojudge.tw/ShowProblem?problemid=c813)