HACKER RANK Project Euler 05 - Smallest Multiple - Solution.'C++'

Project Euler Challenges-05 - Smallest Multiple Solution

Problem Statement

This problem is a programming version of Problem 5 from projecteuler.net

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(divisible with no remainder) by all of the numbers from 1 to N?

Input Format 
First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N.

Output Format 
Print the required answer for each test case.

Constraints 
1≤T≤10 
1≤N≤40

Sample Input

2
3
10

Sample Output

6
2520

Solution:

    #include <iostream>
    using namespace std;
    int a[50][50];
    int main()
    {
     for(int i=2;i<=40;++i)
     {
      int temp=i;
      int j=2;
      while(temp!=1)
      {
       while(temp%j==0)
       {
        temp/=j;
        ++a[i][j];
       }
       ++j;
      }
     }
     int t;
     cin>>t;
     for(int k=1;k<=t;++k)
     {
      int n;
      cin>>n;
      long long ans=1;
      for(int i=2;i<=n;++i)
      {
       int mmax=a[2][i];
       for(int j=2;j<=n;++j)
        mmax=max(mmax,a[j][i]);
       for(int k=1;k<=mmax;++k)
        ans*=i;
      }
      cout<<ans<<endl;
     }
    }

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