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*…AN + A1*A2…*AN-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
数根公式
又因为(x*y)%9 = (x%9)*(y%9)
所以可以推得答案
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define ll long long
int main(){
int n, k;
int a;
int s;
int s1;
scanf("%d", &n);
while(n--){
int i, j;
s=0;
s1 = 1;
scanf("%d", &k);
for(i = 0; i < k; i++){
scanf("%d", &a);
a = a%9;
s1=s1*a%9;
s = (s+s1)%9;
//cout << s << endl;
}
if(s!=0)
printf("%d\n", s);
else
printf("9\n");
}
return 0;
}