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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

using ll=long long;
using int64=int64_t;
const int inf=100000000;

int in(){
  int a;
  cin >> a;
  return a;
}//int型関数の受け取り

void forin(int* x,int y){
  for (int i=0;i<y;i++){
    cin >> x[i];
  }
}//(x,y) int型配列x[y]の受け取り

int intpow(int x,int y){
  double a=x,b=y;
  return pow(a,b);
}//(x,y) xをy乗

int intfabs(int x){
  double a=x;
  return fabs(a);
}//(x) xの絶対値

int fact(int x){
  if (x == 1) return 1;
  return x*fact(x-1);
}//(x) x!

int fib(int x){
  if (x <= 2) return x;
  int a[x];
  a[0]=1;
  a[1]=2;
  for (int i=2;i<x;i++){
    a[i]=a[i-1]+a[i-2];
  }
  return a[x-1];
}//(x) フィボナッチ数列のx番目

int div1(ll x){
  return x%1000000007;
}

int64 gcd(int64 b,int64 a){
    if (b%a==0) return a;
    return gcd(a,b%a);
}

main(){
    int64 a,b,c,d,n,ans=0;//4 9 2 3
    cin >> a >> b >> c >> d;
    if (d<c) swap(c,d);
    if (d%c!=0) n=c*d/gcd(c,d%c);
    else n=d;
    for (int64 i=max(a,n);i<=b;i++){
        if (i%n!=0) ans++;
    }
    cout << ans << endl;
}