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ProjectEulerTS/Problems/Problem21.ts

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TypeScript

//ProjectEulerTS/Problems/Problem21.ts
//Matthew Ellison
// Created: 04-08-21
//Modified: 07-14-21
//Evaluate the sum of all the amicable numbers under 10000
//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/typescriptClasses
/*
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/>.
*/
import { getSum } from "../../../Typescript/typescriptClasses/ArrayAlgorithms";
import { getDivisors } from "../../../Typescript/typescriptClasses/NumberAlgorithms";
import { Problem } from "./Problem";
export class Problem21 extends Problem{
//Variables
//Static variables
private static LIMIT: number = 10000; //The number that is > the largest number to be checked
//Instance variables
private divisorSum: number[]; //Holds the sum of the divisors of the subscript number
private amicable: number[]; //Holds all amicable numbers
//Functions
//Constructor
public constructor(){
super(`Evaluate the sum of all the amicable numbers under ${Problem21.LIMIT}`);
this.divisorSum = [];
this.amicable = [];
this.reserveArray();
}
//Operational functions
//Reserve the size of the array to speed up insertion
private reserveArray(): void{
//Reserving it now makes it faster later
//Make sure the array is filled with 0's
while(this.divisorSum.length < Problem21.LIMIT){
this.divisorSum.push(0);
}
}
//Solve the problem
public solve(): void{
//If the problem has already been solved do nothing and end the function
if(this.solved){
return;
}
//Start the timer
this.timer.start();
//Generate the divisors of all numbers < 10000, get their sum, and add it to the list
for(let cnt = 1;cnt < Problem21.LIMIT;++cnt){
let divisors: number[] = getDivisors(cnt); //Get all the divisors of a number
if(divisors.length > 1){
divisors.pop(); //Remove the last element because it will be the number itself
}
this.divisorSum[cnt] = getSum(divisors);
}
//Check every sum of divisors in the list for matching sum
for(let cnt = 1;cnt < this.divisorSum.length;++cnt){
let sum: number = this.divisorSum[cnt];
//If the sum is greater than the number of divisors then it is impossible to be amicable. Skip the number and continue
if(sum >= this.divisorSum.length){
continue;
}
//We know that divisorSum[cnt] == sum, so if divisorSum[sum] == cnt we found an amicable number
if(this.divisorSum[sum] == cnt){
//A number can't be amicable with itself
if(sum == cnt){
continue;
}
//Add it to the array of amicable numbers
this.amicable.push(cnt);
}
}
//Sort the array for neatness
this.amicable.sort((n1, n2) => n1 - n2);
//Stop the timer
this.timer.stop();
//Throw a flag to show the problem is solved
this.solved = true;
}
//Reset the problem so it can be run again
public reset(): void{
super.reset();
this.divisorSum = [];
this.amicable = [];
this.reserveArray();
}
//Gets
//Returns the result of solving the problem
public getResult(): string{
this.solvedCheck("result");
let result: string = `All amicable numbers less than ${Problem21.LIMIT} are\n`;
for(let cnt = 0;cnt < this.amicable.length;++cnt){
result += `${this.amicable[cnt]}\n`;
}
result += `The sum of all of these amicable numbers is ${getSum(this.amicable)}`;
return result;
}
//Returns a vector with all of the amicable numbers calculated
public getAmicable(): number[]{
this.solvedCheck("amicable numbers");
return this.amicable;
}
//Returns the sum of all of the amicable numbers
public getSum(): number{
this.solvedCheck("sum of the amicable numbers");
return getSum(this.amicable);
}
}
/* Results:
The sum of all of these amicable numbers is 31626
It took an average of 7.454 milliseconds to run this problem through 100 iterations
*/