In this HackerEarth Takeoff problem solution, There are three planes A, B, and C. Plane A will takeoff on every pth day i.e. p, 2p, 3p and so on. Plane B will takeoff on every qth day and plane C will takeoff on every rth day. There is only one runway and the takeoff timing is same for each of the three planes on each day. Your task is to find out the maximum number of flights that will successfully takeoff in the period of N days.


HackerEarth Takeoff problem solution


HackerEarth Takeoff problem solution.

#include <bits/stdc++.h>

using namespace std;

const int N = 1E5 + 5;
int f[N];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t --) {
        int n, p, q, r;
        cin >> n >> p >> q >> r;
        memset(f, 0, sizeof f);
        for(int i = p; i <= n; i += p)
            f[i] ++;
        for(int i = q; i <= n; i += q)
            f[i] ++;
        for(int i = r; i <= n; i += r)
            f[i] ++;
        int cnt = 0;
        for(int i = 1; i <= n; i ++)
            if(f[i] == 1)
                cnt ++;
        cout << cnt << '\n';
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
int main()
    {
        ios_base::sync_with_stdio(0);
        int t;
        cin >> t;
        assert(t >= 1 && t <= 10);
        while(t--){
            int n , p , q , r;
            cin >> n >> p >> q >> r;
            assert(n >= 1 && n <= 100000);
            assert(p >= 1 && p <= 100000);
            assert(q >= 1 && q <= 100000);
            assert(r >= 1 && r <= 100000);
            int idx = p , ans = 0;;
            while(idx <= n){
                if(idx % q != 0 && idx % r != 0){
                    ++ans;
                }
                idx += p;
            }
            idx = q;
            while(idx <= n){
                if(idx % p != 0 && idx % r != 0){
                    ++ans;
                }
                idx += q;
            }
            idx = r;
            while(idx <= n){
                if(idx % q != 0 && idx % p != 0){
                    ++ans;
                }
                idx += r;
            }
            cout << ans << endl;
        }
    }