牛牛的零食（容斥原理）

思路：用一組樣例

5，6，9 區間1——100

答案是 100/8 - 100/lcm(5,8) - 100/lcm(8,6) - 100/lcm(8,9) + 100/lcm(8,6,5) + 100/lcm(8,6,9) + 100/lcm(8,5,9) - 100/lcm(8,5,6,9)

容斥原理，奇數個元素加，偶數個元素減。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=2e6+10;
const int inf=0x7f7f7f7f;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}

ll a[20];
ll n,ans,l,r;
void dfs(int cnt,int pos,ll num)
{
   // cout<<num<<endl;
    if(cnt&1)
        ans-=r/num-l/num;
    else
        ans+=r/num-l/num;
    for(int i=pos+1;i<n;i++)
        dfs(cnt+1,i,lcm(num,a[i]));
}
int main()
{

    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    cin>>l>>r;
    l--;
    dfs(0,-1,8);
    cout<<ans<<endl;





    return 0;
}