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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)








import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Queue;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.Set;



public class Main{
	
	static int posx;
	static int posy;
	static int type;
	static int diff;
	static int INF = Integer.MAX_VALUE;
	static int[][] d;
	
	//最大公約数・最小公倍数（セットで使う）
	public static long gcm(long a, long b) {
		long result = a;
		long k = 0;
		long n = b;
		do {
			k = result % n;
			result = n;
			n = k;
		} while(k != 0);
		return result;
	}
	
	public static long lcm(long a, long b) {
		long g;
		g = gcm(a, b);
		return a/g*b;
	}
	
	static long gcd (long a, long b) {
		long temp;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return b;
	}
	
	//等差数列
	static long tousasum(int n) {
		return (long)n*((long)n+1)/2;
	}



	//素数判定
	static boolean isPrime (int n) {
		if (n==2) return true;
		if (n<2 || n%2==0) return false;
		double d = Math.sqrt(n);
		for (int i=3; i<=d; i+=2) if(n%i==0){return false;}
		return true;
	}


	//倍数判定（10進数以外のときに有用）
	static boolean isMultiple (String s, int base, int m) {
		int temp = 0;
		for (int i=0; i<s.length(); i++) {
			temp = (temp*base+Character.getNumericValue(s.charAt(i)))%m;
		}
		if (temp==0) {return true;}
		return false;
	}


	//階乗
	static long factorial (int i) {
		if (i==1) {return 1;}
		else {return i*factorial(i-1);}
	}


	//進数変換
	static String toNbase (String sm, int m, int n) {
		return Long.toString(Long.parseLong(sm,m),n);
	}
	
	public static final long gcd_n(long[] param) {
	    final int len = param.length;
	    long g = gcd(param[0], param[1]);    //gcd(a, b) は以前作ったもの
	    for (int i = 1; i < len - 1; i++) {
	        g = gcd(g, param[i + 1]);       //gcd(a, b) は以前作ったもの
	    }
	    return g;
	}
	
	public static long lcm_n(long[] numbers) {
		long l;
		l = numbers[0];
		for (int i = 1; i < numbers.length; i++) {
			l = lcm(l, numbers[i]);
		}
		return l;
	}
	
	
 public static void main(String[] args) {
	 Scanner sc=new Scanner(System.in);
	 
	 long A=sc.nextLong();
	 long B=sc.nextLong();
	 long C=sc.nextLong();
	 long D=sc.nextLong();
	 
	 long numC=A/C;
	 if(A%C!=0) {
		 numC+=1;
	 }
	 long numC2=B/C;
	 
	 long numD1=A/D;
	 if(A%D!=0) {
		 numD1+=1;
	 }
	 long numD2=B/D;
	 
	
	 long numCD1=A/lcm(C,D);
	 if(A%(lcm(C,D))!=0) {
		 numCD1+=1;
	 }
	 long numCD2=B/lcm(C,D);
	 
	 
	 
	 
	 long result=(numC2-numC+1)+(numD2-numD1+1)-(numCD2-numCD1+1);
	 
	 //System.out.println(numC+" "+numC2+" "+numD1+" "+numD2+" "+numCD1+" "+numCD2);
	 System.out.println(B-A+1-result); 
	 
	
	 
	 
	 
 }
	


}