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 <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))

/*ユークリッドの互除法を使って、最大公約数を返す関数*/
LL Euqlid_num(LL a, LL b)
{
    LL r;
    r = a % b;
    while(r!=0){
        a = b;
        b = r;
        r = a%b;
    }    
    return b;
}


int main()
{
    LL A, B, C, D;
    LL count;
    LL count_B1, count_B2, count_B3, count_A1, count_A2, count_A3;
    LL least_multi, ans;
    
    
    scanf("%lld %lld %lld %lld", &A, &B, &C, &D);

    /*Bいかに何個の割り切れる数があるかを探す*/
    count_B1 = B / C; count_B2 = B / D;
    least_multi = C * D /(Euqlid_num(C, D));
    count_B3 = B / least_multi;
    count = count_B1 + count_B2 - count_B3;


    /*A以下に何個割り切れる数があるか*/
    count_A1 = (A-1) / C; count_A2 = (A-1) / D;
    count_A3 = (A-1) / least_multi;
    count += count_A3 - count_A1 - count_A2 ;
    ans = B - A - count + 1;
    
    printf("%lld", ans);
    
    return 0;
}