SGU118――Digital Root - c++编程基础

118. Digital Root

time limit per test: 0.5 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

首先，比如样例三个数字，要求是2*3*4+2*3+2，可以这么看2+2*3+2*3*4 还有两个数论公式：（a+b）%c=(a%c)+(b%c) (a*b)%c=(a%c)*(b%c) 还有一个结论是一个数的所有位数的和为这个数对9取模。

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                    using namespace std; int main() { int t,n,a; scanf("%d",&t); while(t--) { int p=1,sum=0; scanf("%d",&n); for(int i=0;i