目录

• Description
• State
• Input
• Output
• Solution
• Code

Description

有 \(n\) 个数，要求找到最小的 \(seed\)， 使得 \(a[i]%seed\) 都不相同

State

\(1<=n<=5*10^5\)

\(0<=a[i]<=5*10^5\)

Input

3
1 2 4

Output

4

Solution

对于任意的 \(i,j\)， \(a[i]!=a[j](mod\ p)\)，也就是 \(a[i]-a[j]!=0(mod\ p)\)，需要计算出所有的差，利用 \(fft/ntt\) 的系数可以快速的解决

得到了所有差值之后，枚举 \(p\)，如果 \(p\) 的倍数存在，那么 \(p\) 是不成立的

---

Code

const int N = 2e6 + 5;
 
    int n, m, k, _;
    //int a[N];
    struct Complex
    {
        double x,y;
        Complex (double x = 0, double y = 0) : x(x), y(y){}
    }a[N],b[N];
Complex operator+(Complex a, Complex b){ return Complex(a.x + b.x ,a.y + b.y); }
Complex operator-(Complex a, Complex b){ return Complex(a.x - b.x ,a.y - b.y); }
Complex operator*(Complex a, Complex b){ return Complex(a.x * b.x - a.y * b.y ,a.x * b.y + a.y * b.x); }

int lim = 1, bit = 0, rev[N];
void init(int n)
{
    while(lim <= n) lim <<= 1, bit ++;
    for(int i = 0; i < lim; i ++){
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    }
}

void fft(Complex *a, int opt)
{
    for(int i = 0; i < lim; i ++){
        if(i < rev[i]) swap(a[i], a[rev[i]]);
    }
    for(int mid = 1; mid < lim; mid *= 2){
        Complex wn(cos(pi / mid), opt * sin(pi / mid));
        for(int len = mid * 2, i = 0; i < lim; i += len){
            Complex w(1, 0);
            for(int j = 0; j < mid; j ++, w = w * wn){
                Complex x = a[i + j], y = a[i + j + mid] * w;
                a[i + j] = x + y;
                a[i + j + mid] = x - y;
            }
        }
    }
    if(opt == 1) return ;
    for(int i = 0; i < lim; i ++){
        a[i].x = int(a[i].x / lim + 0.5);
    }
}

int vis[N];
bool judge(int x)
{
    for(int i = x; i < lim; i += x){
        if(vis[i]) return 0;
    }
    return 1;
}

signed main()
{
    const int _5 = 5e5;
    //IOS;
    while(~ sd(n)){
        init(1e6);
        rep(i, 1, n){
            int x = read();
            a[x] = 1;
            b[_5 - x] = 1;
        }
        fft(a, 1); fft(b, 1);
        for(int i = 0; i < lim; i ++) a[i] = a[i] * b[i];
        fft(a, -1);
        for(int i = 0; i < lim; i ++) vis[abs(i - _5)] = a[i].x;
        for(int i = n;; i ++){
            if(judge(i)){
                pd(i);
                return 0;
            }
        }
        return 0;
    }
    //PAUSE;
    return 0;
}