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<stdlib.h>
#include<math.h>
#include<string.h>

long z(long A,long B);

int main(void){
  long A,B,C,D,i,j,ans,f,e,g,h,k,l,t;
  
  if(scanf("%ld %ld %ld %ld",&A,&B,&C,&D)>4){
    exit(0);
  }
  f=0;
  h=0;
  l=0;
  t = z(C,D);
  //printf("%ld",t);
  e=(B/t)-(A/t);
  f=e;
   if((A%t)<0.1){
     f=f+1;
  }
 if((B%t)<0.1){
   //f=f+1;
  }
  g=(B/C)-(A/C);
  h=g;
  if((B%C)<0.1){
    //h=h+1;
  }
   if((A%C)<0.1){
   h=h+1;
  }
  k=(B/D)-(A/D);
  l=k;
   if((B%D)<0.1){
     //l=l+1;
  }
 if((A%D)<0.1){
   l=l+1;
  }
  ans=(B-A+1)+f-h-l;
  printf("%ld",ans);
  return 0;
}

long z(long A,long B){
  long a,b,x,tmp,r;

  a=A;
  b=B;
  x = a * b;
  
  if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }
 
  /* ユークリッドの互除法 */
  r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }
 
  /* 最小公倍数を出力 */
  return x/b;
}