ai % seed != aj %seed,那么|ai-aj|%seed != 0,|ai-aj|的集合中元素可以用fft组合出来,既然|ai-aj|不能整除seed,同时也说明seed不是任何一个|ai-aj|的因子,因此我们还要筛出每个数的因子,筛因子可以做到O(nlogn)

#include <bits/stdc++.h>

using namespace std;

const int N = 2000010, base = 500001;
const double PI = acos(-1);

struct Complex
{
    double x, y;
    Complex operator+ (const Complex& t)
    {
        return {x + t.x, y + t.y};
    }
    Complex operator- (const Complex& t)
    {
        return {x - t.x, y - t.y};
    }
    Complex operator* (const Complex &t)
    {
        return {x * t.x - y * t.y, x * t.y + y * t.x};
    }
}a[N], b[N];
bool st[N];
int rev[N], tot, bit;

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;
            }
        }
    }
}

bool check(int x)
{
    for (int i = x; i <= base; i += x) 
        if (st[i]) return false;

    return true;
}

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) 
    {
        int w;
        scanf("%d", &w);
        a[w].x = 1;
        b[base - w].x = 1;    //负数做偏移 
    }
    while ((1 << bit) < base * 2) bit ++ ;
    tot = 1 << bit;
    for (int i = 0; i < tot; i ++ )
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    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 = 0; i < tot; i ++ )
    {
        int w = a[i].x / tot + 0.5;
        if (w) st[abs(i - base)] = true;    
    }

    for (int i = n; i <= base; i ++ )
        if (check(i))
        {
            printf("%d\n", i);
            break;    
        } 

    return 0;
}