SPOJ DIVSUM - Divisor Summation

DIVSUM - Divisor Summation

#number-theory

Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.

Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.

e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Input

An integer stating the number of test cases (equal to about 200000), and that many lines follow, each containing one integer between 1 and 500000 inclusive.

Output

One integer each line: the divisor summation of the integer given respectively.

Example

Sample Input:
3
2
10
20

Sample Output:
1
8
22

Warning: large Input/Output data, be careful with certain languages

解题：第一道用RUST写的水题，RUST的效率真心不敢恭维，尤其是IO效率

 use std::io;
 use std::io::prelude::*;
 fn main(){
         let stdin = io::stdin();
         let mut lines = stdin.lock().lines();
         let n = lines.next().unwrap().unwrap().parse::<i32>().unwrap();
         for _ in ..n {
             let mut x = ;
             let mut sum = ;
             let y = lines.next().unwrap().unwrap().parse::<i32>().unwrap();
             while x*x <= y {
                 if y%x == {
                     let tmp = y/x;
                     if x != y {
                         sum += x;
                     }
                     if tmp != x && tmp != y {
                         sum += tmp;
                     }
                 }
                 x = x + ;
             }
             println!("{}",sum);
         }
 }