给定一个数字,找出另外的特定数字的所有唯一倍数的总和,但不包括第一个数字.

如果我们列出20以下，3或5的倍数的所有自然数,我们得到3,5,6,9,10,12,15和18.

这些倍数的总和是78.

在项目Euler中，问题1的变种http://projecteuler.net/problem=1

pub fn sum_of_multiples(limit: u32, factors: &[u32]) -> u32 {
   unimplemented!(
       "Sum the multiples of all of {:?} which are less than {}",
       factors,
       limit
   )
}

# #![allow(unused_variables)]
#fn main() {
#[test]
fn multiples_one() {
   assert_eq!(0, sum_of_multiples(1, &[3, 5]))
}

#[test]
//#[ignore]
fn multiples_two() {
   assert_eq!(3, sum_of_multiples(4, &[3, 5]))
}

#[test]
//#[ignore]
fn multiples_three() {
   assert_eq!(23, sum_of_multiples(10, &[3, 5]))
}

#[test]
//#[ignore]
fn multiples_four() {
   assert_eq!(2318, sum_of_multiples(100, &[3, 5]))
}

#[test]
//#[ignore]
fn multiples_five() {
   assert_eq!(233168, sum_of_multiples(1000, &[3, 5]))
}

#[test]
//#[ignore]
fn multiples_six() {
   assert_eq!(51, sum_of_multiples(20, &[7, 13, 17]))
}

#[test]
//#[ignore]
fn multiples_seven() {
   assert_eq!(30, sum_of_multiples(15, &[4, 6]))
}

#[test]
//#[ignore]
fn multiples_eight() {
   assert_eq!(4419, sum_of_multiples(150, &[5, 6, 8]))
}

#[test]
//#[ignore]
fn multiples_nine() {
   assert_eq!(275, sum_of_multiples(51, &[5, 25]))
}

#[test]
//#[ignore]
fn multiples_ten() {
   assert_eq!(2203160, sum_of_multiples(10000, &[43, 47]))
}

#[test]
//#[ignore]
fn multiples_eleven() {
   assert_eq!(4950, sum_of_multiples(100, &[1]))
}

#[test]
//#[ignore]
fn multiples_twelve() {
   assert_eq!(0, sum_of_multiples(10000, &[]))
}

#}

# #![allow(unused_variables)]
#fn main() {
use std::collections::BTreeSet;

pub fn sum_of_multiples(limit: u32, factors: &[u32]) -> u32 {
   let mut multiples: BTreeSet<u32> = BTreeSet::new();

   for &f in factors {
       let mut multiplier = 2;
       let mut x = f;
       while x < limit {
           multiples.insert(x);
           x = f * multiplier;
           multiplier += 1;
       }
   }

   multiples.iter().sum()
}

#}