[Polynomial algorithm] (Part 2) NTT fast number theory transformation study notes

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\(2.Medium-NTT(FNT)\)

definition

NTT\((Number\ Theoretic\ Transforms)\)

(Also known as \ (Fast \ Number The-Theoretic \ the Transform \) , referred \ (FNT \) )

Chinese name: fast number theory transformation

~~(Not True Transforms)~~

\ (Q: FFT \) understand, \ (NTT \) what use is it? \ (FFT \) has had enough ah? ?

\ (A: \) wait, with respect to \ (an FFT \) is, \ (NTT \) is more targeted, \ (NTT \) may be calculated in polynomial multiplication modulo sense, so as to avoid \ (an FFT \) a \ (Double \) loss of precision of the case, but \ (NTT \) there are special requirements for the modulus, which will be mentioned below.

So \ (NTT \) is how to achieve it?

In fact, \ (NTT \) implementation method and \ (FFT \) is almost the same, just replace the complex arithmetic modulo operation can be.

Now think about why \ (FFT \) point value represents the unit root to use it?

Because the unit has the following properties for the root \ (an FFT \) utilized to accelerate:

\(1.\)

\[\omega_n^n=1\]

This is in \ (DFT \) will need time

\(2.\)

All unit root different from each other, so as to calculate the correct answer (for example, \ (n \) a \ (n \) element equations have a unique solution only linearly independent).

\(3.\)

\[\begin{cases} \omega_{2n}^{2k}=\omega_n^k(k<\frac n2)\\ \omega_n^{k+\frac n2}=-\omega_n^k(k\ge\frac n2) \end{cases}\]

Obviously this partition in order to proceed

\(4.\)

\[\sum_{k=0}^{n-1}\omega_n^{k(j-i)}= \begin{cases} 0(i\not=j)\\ n(i=j) \end{cases}\]

This \ (IDFT \) need time.

So then we need to find the right number to meet several properties instead of the above unit root.

Primitive root

With prime \ (P \) , then \ (P \) primitive root \ (G \) satisfies \ (g ^ k \ mod p (1 \ le k <p-1) \) different from each other.

If the prime number then \ (n-P = Q *. 1 + (n-2 = X ^) \) , there is in accordance with Fermat's little theorem \ (a ^ {p-1 } \ equiv 1 (\ mod p) \)

Obviously, \ (G_0 \ equiv. 1-G_ {P} \ equiv. 1 (\ MOD P) \)

Assuming that \ (\ omega_n = G ^ Q \) , can be obtained \ (n-\) non-identical numbers: \ (\ omega_n ^ K (0 \ Le K <n-) \)

2 properties meet

And \ (\ omega_n ^ n \ equiv g ^ {p-1} \ equiv 1 (\ mod p) \)

Meet Property 1

Then you can get:

\(\because p=q*n+1=\frac q2*2n+1,\omega_n=g^q\)

\(\therefore \omega_{2n}=g^{\frac q2}\)

\(\omega_{2n}^{2k}=g^{\frac q2*2k}=g^{qk}=\omega_n^k\)

And

\(\omega_n^{k+\frac n2}\)

\(=\omega_n^k*\omega_n^{\frac n2}\)

\(\because (\omega_n^{\frac n2})^2=\omega_n^n=1\)

\(\therefore \omega_n^{\frac n2}=\pm1\)

\ (\ because \ omega \) different from each other

\(\therefore \omega_n^{\frac n2}=-1\)

\(\therefore \omega_n^{k+\frac n2}=-\omega_n^k\)

3 properties meet

Last: for \ (\ sum_ {k = 0 } ^ {n-1} \ omega_n ^ {k (ji)} \)

If \ (I = J \) , then:

\(\sum_{k=0}^{n-1}\omega_n^{k(j-i)}\)

\(=n*\omega_n^0\)

\(=n\)

If \ (i \ not = J \) , then:

\(\sum_{k=0}^{n-1}\omega_n^{k(j-i)}\)

\ (= \ FRAC {(n-Q-^. 1). 1 a_0-Q {}} \) (geometric series summation, where \ (Q \) as a common ratio \ ((\ omega_n ^ {ji }) \ ) , \ (a_0 \) led by key \ ((\ omega_n ^ 0) \) )

\(\because q^n-1=\omega_n^{n(j-i)}-1=(\omega_n^n)^{j-i}-1=0\)

\(\therefore \sum_{k=0}^{n-1}\omega_n^{k(j-i)}=0\)

In summary, to meet the nature 4

\ (Q: \) what? How primitive root seek?

\(A:\)~~I know how, Baidu ah~~

Normally only need to remember \ (998,244,353 \) primitive root is \ (3 \) like \ ((998 244 353 * 119 = 2 ^ {23} + 1) \) , look-up table may be: [Miskcoo apos Space] FFT used in various prime number

Thus, according to the above said, \ (NTT \) framework came out:

The complex operation can be replaced by a modulo.

\ (Q: n \) less than \ (2 ^ {23} \) how to do? Fill up too slow.

\ (A: \) The \ (\ omega_n \) into \ (g ^ {\ frac { p-1} {n}} \) can.

P3803 [template] polynomial multiplication (FFT)

\ (Q: \) Why is \ (FFT \) template?

\(A:\)~~NTT's not found~~Because the answer to this question does not exceed \ (998 244 353 \) , and then to the modulo will have no effect.

Code:

#include <cstdio>
#include <cctype>
#include <algorithm>

char In[1000005],*p1=In,*p2=In;
#define Getchar (p1==p2&&(p2=(p1=In)+fread(In,1,1000000,stdin),p1==p2)?EOF:*p1++)
inline int Getint()
{
    register int x=0,c;
    while(!isdigit(c=Getchar));
    for(;isdigit(c);c=Getchar)x=x*10+(c^48);
    return x;
}

const int p=998244353,g=3;

int Pow(int a,int b)
{
    if(b<0)return Pow(Pow(a,p-2),-b);//逆元
    int Res=1;
    for(;b;b>>=1)
    {
        if(b&1)Res=Res*1LL*a%p;
        a=a*1LL*a%p;
    }
    return Res%p;
}

int n,m,Maxl;
int a[2100005],b[2100005];

void NTT(int Pol[],int op)//op=1为DFT,-1为IDFT
{
    for(int i=0,j=0;i<n;++i)
    {
        if(i>j)std::swap(Pol[i],Pol[j]);
        for(int l=n>>1;(j^=l)<l;l>>=1);
    }
    for(register int i=2;i<=n;i<<=1)
    {
        int m=i>>1,Rs=Pow(g,op*(p-1)/i);
        for(register int j=0;j<n;j+=i)
        {
            int Root=1;
            for(register int k=0;k<m;++k)
            {
                int Tmp=Root*1LL*Pol[j+k+m]%p;
                Root=Root*1LL*Rs%p;
                //int Root=Pow(Omega,op*(p-1)/n*k)
                //不预处理单位根，使用秦九韶算法加速
                Pol[j+k+m]=Pol[j+k]-Tmp;
                Pol[j+k]=Pol[j+k]+Tmp;
                Pol[j+k+m]<0?Pol[j+k+m]+=p:0;//减少取模
                Pol[j+k]>=p?Pol[j+k]-=p:0;
            }
        }
    }
}

int main()
{
    n=Getint(),m=Getint();
    for(register int i=0;i<=n;++i)a[i]=Getint();
    for(register int i=0;i<=m;++i)b[i]=Getint();
    for(Maxl=n+m,n=2;n<=Maxl;n<<=1);
    NTT(a,1),NTT(b,1);
    for(int i=0;i<n;++i)a[i]=a[i]*1LL*b[i]%p;
    NTT(a,-1);
    int Invn=Pow(n,p-2);
    for(int i=0;i<=Maxl;++i)printf("%d%c",int(a[i]*1LL*Invn%p),i==n-1?'\n':' ');
    //答案同样/n
    return 0;
}

I believe with \ (FFT \) basis, \ (NTT \) should be very simple.

References :( \ (STO \ Dalao \) )

Fast Fourier Transform from the polynomial multiplication to -Miskcoo

Fourier transform from number theory to transform -Menci