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;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef pair<int,int> pii;
typedef vector< pii > vii;

#define ll          long long
#define ff          first
#define ss          second
#define pb          push_back
#define mkp         make_pair
#define sz          size()

#define forf(a,b,c)    for(int i=a;i<=b;i+=c)
#define forr(a,b,c)    for(int i=a;i>=b;i-=c)
#define all(a)      a.begin(),a.end()
#define mem(a,b)    memset(a,b,sizeof(a))
#define pi          2*acos(0.0)
#define End         return 0;
#define nl endl

#define scan1(a)    scanf("%d",&a)
#define scan2(a,b)  scanf("%d %d",&a,&b)
#define scan3(a,b,c)  scanf("%d %d %d",&a,&b,&c)
#define scanl(a)    scanf("%lld",&a)
#define scanll(a,b) scanf("%lld %lld",&a,&b)
#define scanlll(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define print1(a)    printf("%d",a)
#define print2(a,b)  printf("%d %d",a,b)
#define print3(a,b,c)  printf("%d %d %d",a,b,c)
#define printl(a)    printf("%lld",a)
#define printll(a,b) printf("%lld %lld",a,b)
#define printlll(a,b,c) printf("%lld %lld %lld",a,b,c)

#define MAXN        100000007
#define INF 0x3f3f3f3f
bool mark[100005];vi prime;
void seive(ll N){for(int i=2;i<=N;i++){if(mark[i]==false){prime.pb(i);for(int j=i+i;j<N;j+=i)mark[j]=true;}}}
//freopen("input.txt", "r", stdin);

int main()
{
   ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  ll a,b,c,d;
  cin>>a>>b>>c>>d;
  long long total=b-a+1;
  ll cb=b/c;ll db=b/d;ll ca=--a/c;ll da=a/d;

  ll tc=cb-ca;ll td=(db-da);

  ll lcm=(c/__gcd(c,d))*d;
  ll tla=++a/lcm;ll tlb=b/lcm;ll tl= tlb-tla;
  cout<<total-(tc+td-tl)<<endl;
   return 0;
}