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: Accepted)

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

typedef long long ll;
typedef pair<int,int> P;
typedef map<int,int> M;
typedef vector<int> V;
typedef queue<int> Q;
typedef pair<long,long> PLL;
typedef map<long,long> MLL;
typedef vector<long long> VLL;
typedef vector<pair<int,int>> VP;
typedef vector<vector<int>> VV;
#define INF (int)(1e9+1)
#define MAXX (int)1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define MOD (ll)(1e9+7)
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n-1);i>=0;i--)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)
#define ALL(a)  a.begin(),a.end()
#define RALL(a) a.begin(),a.end(),greater<int>()
#define PRALL(a) a.begin(),a.end(),greater<pair<int,int>>()
#define ROT(a) a.begin(),a.begin()+1,a.end()
#define RROT(a) a.begin(),a.end()-1,a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
//printf("myco\n");//for debug
/*--------------------------------------------*/
ll lcm(ll A, ll B) {//最小公倍数
    ll a=A;
    ll b=B;
    if (a==0 || b==0) {return 0;}
    if (a<0) {a=-a;}
    if (b<0) {b=-b;}
    while(b>0) {
        ll c=a%b;
        a=b;
        b=c;
    }
    return A/a*B;
}
int main(){
    ll a,b,c,d;cin>>a>>b>>c>>d;
    ll e=lcm(c,d);
    a--;
    ll ans=b-a-((b/c-a/c+b/d-a/d)-(b/e-a/e));
    cout<<ans<<endl;
}