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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int main(void){

  long long a,b,c,d,x,y,tmp,sum;
  long long all;
  long long i;
  long long gomi=0;
  long long ans;

  long long gcd,lcm;
  scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
  all=b-a+1;
  /*for(i=a;i<=b;i++){
    if(i%c==0||i%d==0)
    gomi++;

  }
  ans=all-gomi;
  printf("%lld",ans);
  */

  //ｃとｄの最小公倍数を求める。
  x=c;
  y=d;
  if(x<y){
    tmp=x;
    x=y;
    y=tmp;
  }
gcd=y;//ワイが小さいほう
  while(1){

    tmp=x%y;
    if(tmp==0){
      gcd=y;
      break;
    }
    else{
      x=x/y;
      y=tmp;
    }
  }
  lcm=(a*b)/gcd;

long long csum,dsum,sumlcm;

  csum=b/c-a/c;

  dsum=b/d-a/d;

  sumlcm=b/lcm-a/lcm;
  ans=all-(csum+dsum-sumlcm);
  printf("%lld",ans);





  return 0;
}