HackerEarth Numbers II problem solution

In this HackerEarthNumbers II problem solution, we have given two numbers a and b, you have to find the Nth number which is divisible by a or b.

HackerEarth Numbers II problem solution.

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include<stdio.h>

#define mod 1000000007
#define ll long long
#define nax 100005

using namespace std;


long long findTerm(long long a,long long b,long long lcm,long long value)
{

    return (value/a) + (value/b) - (value/lcm);
}


long long  solve(long long a,long long b , long long N)
{


    long long low = 1;
    long long high = 100000000000000000LL;
    long long gcd = __gcd(a,b);
    long long lcm = (a*b)/gcd;
    long long ans;
    while(low<=high)
    {
        long long mid = (low+high)/2;
        long long term=findTerm(a,b,lcm,mid);
        if(term>=N)
        {

            ans = mid;
            high = mid-1;
        }
        else
            low=mid+1;




    }
    return ans;



}

int main()
{


    int t;
    long long a,b,N;
    cin>>t;
    while(t--)
    {

        cin>>a>>b>>N;
        cout<<solve(a,b,N)<<"n";
    }




}

Second solution

#include<bits/stdc++.h>
#define inf 1e18
#define ll long long
using namespace std;
ll a,b,n,l;
ll gcd(ll x,ll y)
{
    if(y==0)return x;
    else return gcd(y,x%y);
}
bool f(ll x)
{
    ll val= x/a + x/b - x/l;
    if(val>=n)return 1;
    return 0;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b;
        cin>>n;
        ll low=min(a,b),high=inf;
        ll ans;
        l=(a*b)/gcd(a,b);
        while(low<=high)
        {
            ll mid=(low+high)/2;
            if(f(mid))
            {
                ans=mid;
                high=mid-1;
            }
            else
                low=mid+1;
        }
        cout<<ans<<"n";
    }
}

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