## SGU 118 解题报告

http://acm.sgu.ru/problem.php?contest=0&problem=118

## 118. Digital Root

time limit per test: 0.50 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*…*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```

`5`

### CODE

```#include<stdio.h>   int main() { int k,n,x,t,sum; scanf("%d",&k); while(k--) { sum=0; t=1; scanf("%d",&n); while(n--) { scanf("%d",&x); x%=9; t*=x; t%=9; sum+=t; sum%=9; } if(sum==0) sum=9; printf("%dn",sum); } return 0; }```
» 本博客采用署名 2.5 中国大陆许可协议进行许可，本文版权归作者所有，欢迎转载，但必须在明显位置给出原文连接。
anyShare分享到：