codeforces 678C C. Joty and Chocolate(水题)

题目链接：

C. Joty and Chocolate

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.

An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.

After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.

Note that she can paint tiles in any order she wants.

Given the required information, find the maximum number of chocolates Joty can get.

 

Input

 

The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).

 

Output

 

Print the only integer s — the maximum number of chocolates Joty can get.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

 

Examples

 

input

5 2 3 12 15

output

39

input

20 2 3 3 5

output

51

题意：

a的倍数给p个巧克力,b的倍数q个巧克力,即是a又是b的倍数都可以,那么[1,n]一共可以得到多少巧克力;

思路：

跟容斥定理差不多,求一个最小公倍数,就好了;

AC代码：

//#include <bits/stdc++.h>

#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef unsigned long long uLL;
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e6+8;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

LL n,a,b,p,q;

LL gcd(LL x,LL y)
{
    if(y==0)return x;
    return gcd(y,x%y);
}
LL lcm(LL x,LL y)
{
    LL temp=gcd(x,y);
    return x/temp*y;
}
int main()
{
    read(n);
    read(a);
    read(b);
    read(p);read(q);
    LL ans=n/a*p+n/b*q-n/lcm(a,b)*min(p,q);
    cout<<ans<<"
";

    return 0;
}