Problem:
Score : 300 points

Problem Statement

You are given four integers A, B, C, and D. Find the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Constraints

1\leq A\leq B\leq 10^{18}

1\leq C,D\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Sample Input 1

4 9 2 3

Sample Output 1

2

5 and 7 satisfy the condition.

Sample Input 2

10 40 6 8

Sample Output 2

23

Sample Input 3

314159265358979323 846264338327950288 419716939 937510582

Sample Output 3

532105071133627368

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <bitset>    // bitset
#include <cctype>    // isupper, islower, isdigit, toupper, tolower
#include <cmath>     // pow
#include <cstdint>   // int64_t, int*_t
#include <cstdio>    // printf
#include <deque>     // deque
#include <iomanip>   // setprecision
#include <iostream>  // cout, endl, cin
#include <map>       // map
#include <queue>     // queue, priority_queue
#include <set>       // set
#include <stack>     // stack
#include <string>    // string, to_string, stoi
#include <tuple>     // tuple, make_tuple
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <utility>       // pair, make_pair
#include <vector>        // vector
using namespace std;

int gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }
// cout << fixed << setprecision(10);
// vector<vector<int>> vec(H, vector<int>(W))    // int H*W
// int binary_search(int key) {
//     int ng = -1;
//     int ok = (int)a.size();
//     while(abs(ok - ng) > 1) {
//         int mid = (ok + ng) / 2;
//         if(isOK(mid, key))
//             ok = mid;
//         else
//             ng = mid;
//     }
//     return ok;
// }
int main() {
    long long A, B, C, D;
    cin >> A >> B >> C >> D;
    long long lcm = C / gcd(C, D) * D;
    cout << B - A + 1 - (B / C - (A - 1) / C) - (B / D - (A - 1) / D) +
                (B / lcm - (A - 1) / lcm)
         << endl;
}