Added solution to problem 23

src/Problems/Problem23.rs

@@ -0,0 +1,99 @@
+//ProjectEulerRust/src/Problems/Problems22.rs
+//Matthew Ellison
+// Created: 06-17-20
+//Modified: 06-17-20
+//What is the total of all the name scores in this file?
+//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/RustClasses
+/*
+	Copyright (C) 2020  Matthew Ellison
+
+	This program is free software: you can redistribute it and/or modify
+	it under the terms of the GNU Lesser General Public License as published by
+	the Free Software Foundation, either version 3 of the License, or
+	(at your option) any later version.
+
+	This program is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU Lesser General Public License for more details.
+
+	You should have received a copy of the GNU Lesser General Public License
+	along with this program.  If not, see <https://www.gnu.org/licenses/>.
+*/
+
+
+extern crate myClasses;
+use crate::Problems::Answer::Answer;
+
+pub fn getDescription() -> String{
+	"What is the total of all the name scores in this file?".to_string()
+}
+
+pub fn solve() -> Answer{
+	let MAX_NUM = 28123;
+	//Setup the variables
+	let mut divisorSums = Vec::<i64>::new();	//Holds the sum of all the divisors of a number
+	//Make sure every element has a 0 in it's location
+	for _ in 0..=MAX_NUM{
+		divisorSums.push(0);
+	}
+
+	//Start the timer
+	let mut timer = myClasses::Stopwatch::Stopwatch::new();
+	timer.start();
+
+	//Get the sum of the divisors of all numbers < MAX_NUM
+	for cnt in 1..MAX_NUM{
+		let mut div = myClasses::Algorithms::getDivisors(cnt);
+		//Remove the last element, which is the number itself. This gives us the propper divisors
+		if(div.len() > 1){
+			div.remove(div.len() - 1);
+		}
+		divisorSums[cnt as usize] = div.iter().sum();
+	}
+
+	//Get the abundant numbers
+	let mut abund = Vec::<i64>::new();
+	for cnt in 0..divisorSums.len(){
+		if(divisorSums[cnt] > cnt as i64){
+			abund.push(cnt as i64);
+		}
+	}
+
+	//Check if each number can be the sum of 2 abundant numbers and add to the sum if no
+	let mut sum = 0;
+	for cnt in 1..MAX_NUM{
+		if(!isSum(&abund, cnt)){
+			sum += cnt;
+		}
+	}
+
+	//Stop the timer
+	timer.stop();
+
+	//Return the result
+	return Answer::new(format!("The answer is {}", sum), timer.getString());
+}
+
+fn isSum(abund: &Vec<i64>, num: i64) -> bool{
+	//Pick a number for the first part of the sum
+	for firstNum in 0..abund.len(){
+		//Pick a number for the second part of the sum
+		for secondNum in firstNum..abund.len(){
+			let sum = abund[firstNum] + abund[secondNum];
+			if(sum == num){
+				return true;
+			}
+			else if(sum > num){
+				break;
+			}
+		}
+	}
+	//If you have run through the entire list and did not find a sum then it is false
+	return false;
+}
+
+/* Results:
+The answer is 4179871
+It took 4.902 seconds to solve this problem
+*/