Project Euler: Solution for Problem 5

Smallest multiple

Problem 5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

 

Source link: https://projecteuler.net/problem=5 

 

Solution: implemented in C++.

This problem could be simplified to two question: 

1. All prime number cannot be divide, thus we could mutiple all prime numbers first.
2. Based on the prouction of prime numbers we got, we can calculate the lowest common multiple number of composite numbers and production we already have, if only the production result is not evenly divisible by the composite number.

 

#include<iostream>
#include<string>
#include<vector>
#include<cmath>
using namespace std;
    
int fac(int origin, int comps, vector<int> prs);

void main()
{
    const int num = 20;　　
    int pr_product = 1;　　//the result stores in this varaiable
    vector<int> pr;　　//vector containing all prime numbers in range (0,20)
    vector<int> compo;　　//vector containing all composite numbers in range (0,20)
    //find out all prime numbers among 1-20
    for(int i = 2; i < (num + 1); i++){
        bool isPr = true;
        for(int cnt = 2; cnt < i; cnt++){
            if( i%cnt == 0){
                isPr = false;
                break;
            }
        }
        if(isPr){　　//prime numbers push into vector pr
            pr_product *= i;
            pr.push_back(i);
        }else{　　　　//composite numbers push into vector compo
            compo.push_back(i);
        }
    }
    for(int i = 0; i < compo.size(); i++){　　//For each composite number which the result we have now cannot divide
        if(pr_product%compo[i] != 0)　　　　　　//determine lowest common multiple
            pr_product = fac(pr_product, compo[i], pr);
    }
    cout<<pr_product<<endl;
}

int fac(int origin, int comps, vector<int> prs){　　//determine lowest common multiple
    int record = 1;
    for(int i = 0; comps > prs[i]; i++){
        while( (origin%prs[i] == 0) && (comps%prs[i] == 0) ){
            record *= prs[i];
            origin /= prs[i];
            comps /= prs[i];
        }
    }
    return record = origin * comps * record;
}