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Last active June 8, 2018 01:14
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Constructing a Number - HackerRank Problems: https://www.hackerrank.com/challenges/constructing-a-number
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#![feature(iterator_step_by)]
use std::io::{self, Read};
/**
* Divisibility by 3 or 9.
* - First, take any number (for this example it will be 492)
* and add together each digit in the number (4 + 9 + 2 = 15).
* Then take that sum (15) and determine if it is divisible by 3.
* - The original number is divisible by 3 (or 9)
* if and only if the sum of its digits is divisible by 3 (or 9).
*/
fn can_construct(nums: &mut [u32]) -> &'static str {
let sum: u32 = nums.iter().sum();
if sum % 3 == 0 {
"Yes"
} else {
"No"
}
}
fn main() {
let mut line = Vec::new();
let stdin = io::stdin();
stdin
.lock()
.read_to_end(&mut line)
.expect("Could not read line");
let buf = String::from_utf8(line).expect("Error While Convert STDIN to String");
for l in buf.lines().skip(2).step_by(2) {
let str_nums: Vec<&str> = l.split_whitespace().collect();
let mut nums: Vec<u32> = str_nums
.into_iter()
.map(|num| num.parse().expect("Error While Converting To u32"))
.collect();
println!("{}", can_construct(&mut nums));
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_can_construct() {
assert_eq!(can_construct(&mut [1, 3, 5]), "Yes"); // 135
assert_eq!(can_construct(&mut [2, 3, 5, 2]), "Yes"); // 2352
assert_eq!(can_construct(&mut [90, 40, 50]), "Yes"); // 904050
assert_eq!(can_construct(&mut [1999, 2000, 2001, 2003]), "No"); // 1999200020012003
}
}
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