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;

#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
#define Eunique(v) v.erace(unique(all(v)),v.end())
#define pvec(v)　rep(i,v.size()){cout << v[i] << " ";}
#define INF 0x3f3f3f3f
#define mset(m,v) memset(m,v,sezeof(m))
#define mod (int)(1e9 +7);

const double pi = acos(-1.0);

typedef vector<int> vi;
typedef string ss;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef deque<ll> dll;
typedef pair<ll,ll> P;
ll dx[4] = {1,0,-1,0},dy[4] = {0,1,0,-1};

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

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b; // Be careful not to overflow
}

vll binary_convert(ll n){
    vll s;
    ll size = 0;

    while(n / (ll)(pow(2,size)) > 0)size++;
    rep(i,size + 1){
        s.pb(n / (ll)(pow(2,i)));
        n %= (ll)(pow(2,i));
    }

    return s;
}

ll fib(ll n){
    vll a(n+1);
    a[0] = 1;
    a[1]  = 1;

    REP(i,2,n+1){
        a[i] = a[i-1] + a[i-2];
        a[i] %= mod;
    }

    return a[n-1];
}

int main()
{
    ll a,b,c,d;
    cin >> a >> b >> c >> d;

    ll tmp1 = b/c - a/c;
    ll tmp2 = b/d - a/d;
    ll tmp3 = b/lcm(c,d) - a/(lcm(c,d));

    ll tmp4 = tmp1 + tmp2 - tmp3;
    if(a % c == 0)tmp4++;

    cout << b - a  - tmp4 + 1 << endl;
    return 0;
}