The project is for euler-algorithm resolved code write by rust, as personal algorithm learning project In my code, I will show the problem, and my personal problem solving
/*
Multiples of 3 and 5
- Problem 1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
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** 解:
[1 - 1000)内的倍数之和, 可以用等差数列的方式的求和方法来处理, 3和5是两个等差数列, 最后Sn_3 + Sn_5
*/
pub fn entry() {
let range_max = 1000;
let multiple_3 = 3;
let multiple_5 = 5;
let sum = sum_multiple(multiple_3, range_max) + sum_multiple(multiple_5, range_max);
assert_eq!(sum, test(multiple_3, multiple_5, range_max));
println!("sum of all the multiples of 3 or 5 below 1000, result is {}", sum);
}
#[warn(dead_code)]
fn sum_multiple(multiple: u32, in_range_max: u32) -> u32 {
let n = (in_range_max - 1) / multiple; // - 1, cause [1, 1000)
let a_1 = 1 * multiple;
let a_n = n * multiple;
(a_1 + a_n) / 2 * n // Sn = ((a1 + an) / 2) * n
}improve together