Sgu 118

118. Digital Root

time limit per test: 0.25 sec. memory limit per test: 4096 KB

 

Let f(n) be a sum of digits for positive integer n. If f(n) is one-digit number then it is a digital root for n and otherwise digital root of n is equal to digital root of f(n). For example, digital root of 987 is 6. Your task is to find digital root for expression A₁*A₂*…*A_N + A₁*A₂*…*A_N-1 + … + A₁*A₂ + A₁.

 

Input

Input file consists of few test cases. There is K (1<=K<=5) in the first line of input. Each test case is a line. Positive integer number N is written on the first place of test case (N<=1000). After it there are N positive integer numbers (sequence A). Each of this numbers is non-negative and not more than 10⁹.

 

Output

Write one line for every test case. On each line write digital root for given expression.

 

Sample Input

1
3 2 3 4

Sample Output

5

---

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff 
#define mod 1000000007
#define ll long long
#define maxn 20025
#define pi acos(-1.0)  
int n, m, k, c,t;
int main(){
    scanf("%d", &t);
    while (t--){
        scanf("%d", &n);
        int s = 1, ans = 0, x;
        for (int i = 0; i < n; i++){
            scanf("%d", &x);
            s = s*(x%9) % 9;
            ans = (ans + s) % 9;
        }
        if (ans == 0)ans = 9;
        printf("%d\n", ans);
    }
    return 0;
}