SGU118——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 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 off(n). For example, digital root of 987 is 6. Your task is to find digital root for expressionA₁*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

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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。

 

#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<map>#include<deque>#include<list>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<n;i++)        {            scanf("%d",&a);            a%=9;            p*=a;            p%=9;            sum+=p;        }        sum%=9;        if(sum==0)        printf("9\n");        else        printf("%d\n",sum);    }    return 0;}