https://ac.nowcoder.com/acm/contest/11166/H

 官方题解。

比赛时，我们都是用暴力写的，数据太弱了，今天突然想起来，用fft写了一下。

主要使用fft求差值的卷，乘法我们在多项式的系数上做，加减法 我们在多项式的指数上做。

做减法的时候，我们把负数加N，变成正数即可。

下面是代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <complex>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define sc secondusing namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<ll,ll> pii;
const ll mod=1e9+7;
const ll N =2e6+10;
const double eps = 1e-6;
const double PI=acos(-1);
ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
int dx[8]= {1,0,-1,0,1,1,-1,-1}, dy[8] = {0,1,0,-1,1,-1,1,-1};
struct Complex
{double x, y;Complex operator+ (const Complex& t) const{return {x + t.x, y + t.y};}Complex operator- (const Complex& t) const{return {x - t.x, y - t.y};}Complex operator* (const Complex& t) const{return {x * t.x - y * t.y, x * t.y + y * t.x};}
}a[N], b[N];
int rev[N], bit, tot;
void fft(Complex a[], int inv)
{for (int i = 0; i < tot; i ++ )if (i < rev[i])swap(a[i], a[rev[i]]);for (int mid = 1; mid < tot; mid <<= 1){auto w1 = Complex({cos(PI / mid), inv * sin(PI / mid)});for (int i = 0; i < tot; i += mid * 2){auto wk = Complex({1, 0});for (int j = 0; j < mid; j ++, wk = wk * w1){auto x = a[i + j], y = wk * a[i + j + mid];a[i + j] = x + y, a[i + j + mid] = x - y;}}}
}
void init(){tot = 1 << bit;for (int i = 0; i < tot; i ++ )rev[i] = ((rev[i >> 1] >> 1)) | ((i & 1) << (bit - 1));
}
int n,m=500000;
void solve(){cin>>n;for(int i=1;i<=n;i++){int u;cin>>u;a[u].x=1;b[500000-u].x=1;}while((1<<bit)<2*m+1) bit++;init();fft(a,1);fft(b,1);for(int i=0;i<tot;i++) a[i]=a[i]*b[i];fft(a,-1);// for(int i=500000;i>=500000-5;i--) cout<<(int)((a[i].x/tot)+0.5)<<endl;for(int i=n;i<=500001;i++){int f=1;for(int j=i;j<=500001;j+=i) if((int)((a[500000-j].x/tot)+0.5)>0){f=0;break;}if(f){cout<<i<<"\n";break;}}
}int main()
{iosint T=1;//cin>>T;//T=1;while(T--){solve();}return 0;
}