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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
	long long A, B, C, D, x, y, CD, ret,calc = 1,i = 2,tmpC,tmpD,cnt = 0;
  	long long arr[3];
  	scanf("%lld %lld %lld %lld\n",&A,&B,&C,&D);
  	tmpC = C;
  	tmpD = D;
  	/*if(A > B){
    	x = A;
      	y = B;
    }else{
    	x = B;
      	y = A;
    }*/
  
  	while(cnt <= D){
    	if(tmpC % i == 0 && tmpD % i == 0){
         	tmpC /= i;
          	tmpD /= i;
        	calc *= i;
          	//printf("%lld\n",i);
          	i = 2;
        }else{
      		i++;
        }
      	cnt++;
    }
  	
  	CD = C*D/calc;
  	//printf("%lld\n",CD);
  
  	arr[0] = B/C - A/C;
  	if(A%C == 0){
    	arr[0] += 1;
    }
  	arr[1] = B/D - A/D;
  	if(A%D == 0){
    	arr[1] += 1;
    }
  	if(CD != 1){
  		arr[2] = B/CD - A/CD;
      	if(A%CD == 0){
        	arr[2] += 1;
        }
    }else{
    	arr[2] = 0;
    }
  
  	//printf("arr[0]:%lld arr[1]:%lld arr[2]:%lld\n",arr[0],arr[1],arr[2]);
	
	ret = arr[0] + arr[1] - arr[2];
  	printf("%lld\n",B-A-ret + 1);
  
	return 0;
}