CF-1183F

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题意

给 $n$ 个数，选至多 $3$ 个，且两两不为倍数关系，问最大值是多少

题解

排序，若最大值不取 $a_n$，那只有一种可能 $\frac 1 2 a_n ,\frac 1 3 a_n,\frac 1 5 a_n $，因为如果最大值包含不为 a_n 的因子，那么显然取 a_n 更优。易证，若取 $a_n$ ，一定会取 不为 $a_n$ 因子的最大值，

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }
 
int a[N];
unordered_set<int>s;
int main() {
    int q,n;
    in(q);
    while(q--){
        in(n);
        for0(i,n){
            in(a[i]);
            s.insert(a[i]);
        }
        sort(a,a+n);
        n=unique(a,a+n)-a;
        int maxx=a[n-1];
        int tmp=maxx;
        if(maxx%2==0&&maxx%3==0&&maxx%5==0&&s.count(maxx/2)&&s.count(maxx/3)&&s.count(maxx/5)){
            maxx=maxx/30*31;
        }
        for(int i=n-2;i>=0;i--){
            if(tmp%a[i]){
                for(int j=i-1;j>=0;j--){
                    if(tmp%a[j]&&a[i]%a[j]){
                        tmp+=a[j];
                        break;
                    }
                    
                }
                tmp+=a[i];
                break;
            }
        }
        outln(max(maxx,tmp));
        s.clear();
    }
    return 0;
}