NTT模板

 

NTT(快速数论变换)用到的各种素数及原根：

https://blog.csdn.net/hnust_xx/article/details/76572828

 

NTT多项式乘法模板

#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

typedef long long LL;

const LL mod=998244353; //119*2^23+1   g=3

const int N=(1<<19)+2;

const int g=3;

int rev[N];

LL a[N],b[N];

template<typename T>
void read(T &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
} 

LL Pow(LL a,LL b)
{
    LL res=1;
    for(;b;a=a*a%mod,b>>=1)
        if(b&1) res=res*a%mod;
    return res;
}

void NTT(LL *a,int n,int f)
{
    for(int i=0;i<n;++i)
        if(i<rev[i]) swap(a[i],a[rev[i]]);
    for(int i=1;i<n;i<<=1)
    {
        LL wn=Pow(g,(mod-1)/(i*2));
        if(f<0) wn=Pow(wn,mod-2);
        for(int j=0,p=i<<1;j<n;j+=p)
        {
            LL w=1;
            for(int k=0;k<i;++k,w=w*wn%mod)
            {
                int x=a[j+k],y=w*a[j+k+i]%mod;
                a[j+k]=(x+y)%mod; a[j+k+i]=(x-y+mod)%mod;
            }
        }
    }
    if(f<0)
    {
        LL inv=Pow(n,mod-2);
        for(int i=0;i<n;++i) a[i]=a[i]*inv%mod;
    }
}

int main()
{
    int nn,mm; 
    read(nn); read(mm);
    for(int i=0;i<=nn;++i) read(a[i]);
    for(int i=0;i<=mm;++i) read(b[i]);
    int l=0,len=nn+mm;
    int n;
    for(n=1;n<=len;n<<=1) l++;
    for(int i=0;i<n;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<l-1);
    NTT(a,n,1);
    NTT(b,n,1);
    for(int i=0;i<n;++i) a[i]=a[i]*b[i]%mod;
    NTT(a,n,-1);
    for(int i=0;i<=len;++i) printf("%lld ",a[i]);
}

 

来源：oschina

链接：https://my.oschina.net/u/4383037/blog/4033413