Added solution to problem 33

src/Problems.rs

@@ -55,13 +55,14 @@ pub mod Problem29;
 pub mod Problem30;
 pub mod Problem31;
 pub mod Problem32;
+pub mod Problem33;
 pub mod Problem67;
 
 
-pub static problemNumbers: [u32; 34] = [ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9,
+pub static problemNumbers: [u32; 35] = [ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9,
 									10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
 									20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
-									30, 31, 32, 67];
+									30, 31, 32, 33, 67];
 pub static tooLong: [u32; 7] = [3, 5, 15, 23, 24, 25, 27];
 
 pub fn solveProblem(problemNumber: u32, description: bool, solve: bool) -> Answer::Answer{
@@ -326,6 +327,14 @@ pub fn solveProblem(problemNumber: u32, description: bool, solve: bool) -> Answe
 			answer = Problem32::solve();
 		}
 	}
+	else if(problemNumber == 33){
+		if(description){
+			println!("{}", Problem33::getDescription());
+		}
+		if(solve){
+			answer = Problem33::solve();
+		}
+	}
 	else if(problemNumber == 67){
 		if(description){
 			println!("{}", Problem67::getDescription());

src/Problems/Problem33.rs

@@ -0,0 +1,99 @@
+//ProjectEulerRust/src/Problems/Problems33.rs
+//Matthew Ellison
+// Created: 02-07-21
+//Modified: 02-07-21
+/*
+The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s
+We shall consider fractions like, 30/50 = 3/5, to be trivial examples
+There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator
+If the product of these four fractions is given in its lowest common terms, find the value of the denominator
+*/
+//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/RustClasses
+/*
+	Copyright (C) 2021  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{
+	"If the product of these four fractions is given in its lowest common terms, find the value of the denominator".to_string()
+}
+
+pub fn solve() -> Answer{
+	let mut numerators = Vec::<i32>::new();
+	let mut denominators = Vec::<i32>::new();
+
+	//Start the timer
+	let mut timer = myClasses::Stopwatch::Stopwatch::new();
+	timer.start();
+
+	for denominator in 11..=99{
+		for numerator in 10..denominator{
+			let num1 = numerator.to_string();
+			let denom = denominator.to_string();
+			let mut tempNum = 0;
+			let mut tempDenom = 1;
+
+			//Check that this isn't a trivial example
+			if((num1.chars().nth(1).unwrap() == '0') && (denom.chars().nth(1).unwrap() == '0')){
+				continue;
+			}
+			//Remove the offending digits if they exist
+			else if(num1.chars().nth(0).unwrap() == denom.chars().nth(0).unwrap()){
+				tempNum = num1.chars().nth(1).unwrap() as i32 - 48;
+				tempDenom = denom.chars().nth(1).unwrap() as i32 - 48;
+			}
+			else if(num1.chars().nth(0).unwrap() == denom.chars().nth(1).unwrap()){
+				tempNum = num1.chars().nth(1).unwrap() as i32 - 48;
+				tempDenom = denom.chars().nth(0).unwrap() as i32 - 48;
+			}
+			else if(num1.chars().nth(1).unwrap() == denom.chars().nth(0).unwrap()){
+				tempNum = num1.chars().nth(0).unwrap() as i32 - 48;
+				tempDenom = denom.chars().nth(1).unwrap() as i32 - 48;
+			}
+			else if(num1.chars().nth(1).unwrap() == denom.chars().nth(1).unwrap()){
+				tempNum = num1.chars().nth(0).unwrap() as i32 - 48;
+				tempDenom = denom.chars().nth(0).unwrap() as i32 - 48;
+			}
+
+			//Test if the new fraction is the same as the old one
+			if((tempNum as f64 / tempDenom as f64) == (numerator as f64 / denominator as f64)){
+				numerators.push(numerator);
+				denominators.push(denominator);
+			}
+		}
+	}
+
+	//Get the product of the numbers
+	let numProd = numerators.iter().product::<i32>();
+	let denomProd = denominators.iter().product::<i32>();
+	//Get the gcd to reduce to lowest terms
+	let gcd = myClasses::Algorithms::gcd(numProd, denomProd);
+	//Save the denominator
+	let prodDenominator = denomProd / gcd;
+
+	//Stop the timer
+	timer.stop();
+
+	return Answer::new(format!("The denominator of the product is {}", prodDenominator), timer.getString(), timer.getNano());
+}
+
+/* Results:
+The denominator of the product is 100
+It took an average of 717.868 microseconds to run this problem through 100 iterations
+*/