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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
long long int CD(long long int a,long long int b){
	long long int tmp,r;
	long long int x;
	x=a*b;
	if(a<b)
	{
		tmp = a;
		a = b;
		b = tmp;
	}
	r = a % b;
	while(r!=0)
	{
		a = b;
		b = r;
		r = a % b;
	}
	return x/b;
}
int main()
{
	long long int A,B,C,D,cd,N,cn,dn,cdn;
	scanf("%lld %lld %lld %lld",&A,&B,&C,&D);
	N=B-A+1;
	if(A%C==0){cn = A;}else{cn = A+C-A%C;}
	if(A%D==0){dn = A;}else{dn = A+D-A%D;}
	if(cn>B){cn=0;}else{cn = (B-cn)/C+1;}
	if(dn>B){dn=0;}else{dn = (B-dn)/D+1;}
	cd = CD(C,D);
	if(A%cd==0){cdn = A;}else{cdn = A+cd-A%cd;}
	if(cdn>B){cdn=0;}else{cdn = (B-cdn)/cd+1;}
	printf("%lld\n",N-cn-dn+cdn);
	return 0;
}