sgu 118 Digital Root

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 integern. 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 of987 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 numberN is written on the first place of test case (N<=1000). After it there areN positive integer numbers (sequence A). Each of this numbers is non-negative and not more than10⁹.

Output

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

Sample Input

13 2 3 4

Sample Output

5题目不是很难，很容易知道结论就是模9但是要注意如果整除9的话答案是9不是0哦代码：
#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))using namespace std;int const nMax = 40000;typedef long long LL;typedef pair<LL,LL> pij;LL k,n,a,ans,b;int main(){    scanf("%lld",&k);    while(k--){        scanf("%lld",&n);        ans=0;        b=1;        for(int i=0;i<n;i++){            scanf("%lld",&a);            a%=9;            b*=a;            b%=9;            ans+=b;            ans%=9;        }        if(ans==0)ans=9;        printf("%lld\n",ans);    }    return 0;}