Nth Magical Number in C++

A number is said to be a magical number, if that number is divisible by A or B. We have to find the Nth magical number. As the answer may very large, we will return it modulo 10^9 + 7.

So, if the input is like N = 4, A = 4, B = 3, then the output will be 8

To solve this, we will follow these steps −

• Define a function cnt(), this will take x, A, B,

• return (x / A) + (x / B) - (x / lcm of A and B)

• From the main method, do the following −

• l := 2, r := 1^14, ret := 0

• while l <= r, do −

  • mid := l + (r - l) / 2

  • k := cnt(mid, A, B)

  • if k < N, then −

    • l := mid + 1

  • otherwise −

    • ret := mid

    • r := mid - 1

• ret := ret mod (10^9 + 7)

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; typedef long long int lli; const lli MOD = 1e9 + 7; class Solution {    public:    int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }    int lcm(int a, int b) { return (a * b) / gcd(a, b); }    lli cnt(lli x, lli A, lli B) {       return (x / A) + (x / B) - (x / lcm(A, B));    }    int nthMagicalNumber(int N, int A, int B) {       lli l = 2;       lli r = 1e14;       lli ret = 0;       while (l <= r) {          lli mid = l + (r - l) / 2;          lli k = cnt(mid, A, B);          if (k < N) {             l = mid + 1;          } else {             ret = mid;             r = mid - 1;          }       }       ret %= MOD;       return ret;    } }; main(){    Solution ob;    cout << (ob.nthMagicalNumber(4, 4, 3)); }

Input

4, 4, 3

Output

8