SPOJ DIVSUM

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 0..n {
            let mut x = 1;
            let mut sum = 0;
            let y = lines.next().unwrap().unwrap().parse::<i32>().unwrap();
            while x*x <= y {
                if y%x == 0{
                    let tmp = y/x;
                    if x != y {
                        sum += x;
                    }
                    if tmp != x && tmp != y {
                        sum += tmp;
                    }
                }
                x = x + 1;
            }
            println!("{}",sum);
        }
}