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;
}
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