设这个素数为N,N-1必须要可以被小于n的2的次方整除,因为经常需要除以2的次方长度,必须要可以被整除,所以N可以表示为m*2^k+1的长度,后面我试了一下其他2^k>=ntt长度的素数,发现结果并不正确,后来才发现N还必须大于最大卷积结果N*max{a[i]}*{b[i]},否则mod就会产生错误.后面我又看了一下本原元的存在性证明,那个大佬用一句话就证明了,说是什么群论里的基础知识.没看懂顿时觉得自己对数论很小白.感觉编码课上好像学过什么本原元的,但是没有要证明什么的.好像只有一些结论没有过程.
不过我觉得还是FFT适用的场景更多,虽然精度上有一点问题.
贴一个两三个月前写的NTT题,路过需要的可以参考一下(hihocoder 1388,内含fft板子)
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#pragma warning(disable:4996)
using namespace std;
const int g = 3;
const long long int mymod = 31525197391593473LL;
int revg;
const double pi = acos(-1.0);
const int bign = 100033;
struct complex
{
double r, v;
complex()
{
r = 0.0;
v = 0.0;
}
complex(double r1, double v1)
{
r = r1;
v = v1;
}
complex operator+(const complex& ano)
{
return complex(r + ano.r, v + ano.v);
}
complex operator-(const complex& ano)
{
return complex(r - ano.r, v - ano.v);
}
complex operator*(const complex& ano)
{
return complex(r * ano.r - v * ano.v, r * ano.v + v * ano.r);
}
complex operator/(double r1)
{
return complex(r / r1, v / r1);
}
}a[4*bign],b[4*bign];
long long int mymul(long long int ta, long long int tb)
{
long long int res = 0;
for (;tb; tb >>= 1)
{
if (tb & 1)
res = (res + ta) % mymod;
ta = (ta << 1) % mymod;
}
return res;
}
long long mul(long long x, long long y) {
return (x * y - (long long)(x / (long double)mymod * y + 1e-3) * mymod + mymod) % mymod;
}
long long int quickpow(long long ta, long long int tb)
{
long long res = 1;
for (; tb; tb >>= 1)
{
if (tb & 1)
res = mul(res, ta);
ta = mul(ta, ta);
}
return res;
}
int rev(int num, int len)
{
int ans = 0;
for (int i = 1,j = 0; j<len; i<<=1,j++)
{
if (i & num)
{
ans += (1<<(len - 1 - j));
}
}
return ans;
}
void NTT(long long int c[], int len, int on)
{
int tlen = (1 << len);
for (int i = 0; i < tlen; i++)
{
int j = rev(i, len);
if (j > i)
swap(c[i], c[j]);
}
//long long int tg = g;
//if (on == -1)
// tg = quickpow(g, mymod - 2);
for (int i = 2; i <= tlen; i <<= 1)
{
long long int wn = quickpow(g,(mymod-1)/i);
if (on == -1)
wn = quickpow(wn, mymod-2);
for (int j = 0; j < tlen; j += i)
{
long long int w = 1;
for (int k = j; k < j + i / 2; k++)
{
long long int u = c[k];
long long int v = mul(w , c[k + i / 2]);
c[k] = (u + v) % mymod;
c[k + i / 2] = (u - v + mymod) % mymod;
// long long int oldw = w;
w = mul(w, wn);
}
}
}
if (on == -1)
{
//for (int i = 0; i < tlen / 2; i++)
//{
// swap(c[i], c[tlen - i - 1]);
//}
for (int i = 0; i < tlen; i++)
c[i] = mul(c[i], quickpow(tlen, mymod - 2));
}
}
void FFT(complex c[], int len, int on)
{
int tlen = (1 << len);
for (int i = 0; i < tlen; i++)
{
int j = rev(i,len);
if (j > i)
swap(c[i], c[j]);
}
for (int i = 2; i <= tlen; i <<= 1)
{
complex wn(cos(2*pi/i),on * sin(2*pi/i));
for (int j = 0; j < tlen; j += i)
{
complex w(1,0);
for (int k = j; k < j + i/2; k++)
{
complex u = c[k];
complex v = (w * c[k + i / 2]);
c[k] = u + v;
c[k + i / 2] = u - v;
w = w * wn;
}
}
}
if (on == -1)
{
for (int i = 0; i < tlen; i++)
{
c[i].r /= tlen;
c[i].v /= tlen;
}
}
}
long long int a1[4 * bign], b1[4 * bign];
int main()
{
int T;
//printf("%lld\n", quickpow(g, mymod - 1));
scanf("%d", &T);
while (T--)
{
int n;
long long int ans = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
int x;
scanf("%d", &x);
ans += 1ll * x * x;
a1[i] = x;
a1[i + n] = a1[i];
}
for (int i = n - 1; i >= 0; i--)
{
int x;
scanf("%d", &x);
ans += 1ll * x * x;
b1[i] = x;
}
int len = 0;
int j;
for (j = 1; j < 2 * n; j <<= 1,len++);
for (int i = n; i < j; i++)
b1[i] = 0;
for (int i = 2 * n; i < j; i++)
a1[i] = 0;
NTT(a1, len, 1);
NTT(b1, len, 1);
for (int i = 0; i < j; i++)
a1[i] = mul(a1[i],b1[i])%mymod;
NTT(a1, len, -1);
long long int tmax = 0;
for (int i = n - 1; i < 2 * n - 1; i++)
{
long long int tmp = a1[i];
if (tmp > tmax)
tmax = tmp;
}
printf("%lld\n", ans - 2* tmax);
}
}
/*
5
5
1000000 1000001 1000002 1000003 1000004
1000003 1000004 1000000 1000001 1000002
*/