P3803_NTT

NTT的做法

---

ec好像是linux的机子，回头学下gdb调试

#include <cstdio>
#include <iostream>
#include <cmath>

typedef long long ll;
const int NR = 1 << 22, g = 3, gi = 332748118, mod = 998244353;

using namespace std;
int n, m, rev[NR];
ll a[NR], b[NR];
ll quikpow(ll a, ll k)
{
    ll res = 1;
    while (k > 0)
    {
        if (k&1)res=res*a%mod;
        a=a*a%mod;
        k>>=1;
    }
    return res;
}

void NTT(ll *a,int n,int type){
    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 gn=quikpow(type?g:gi,(mod-1)/(i<<1));
        for(int j=0;j<n;j+=(i<<1)){
            ll g0=1;
            for(int k=0;k<i;++k,g0=g0*gn%mod){
                ll x=a[j+k],y=g0*a[i+j+k]%mod;
                a[j+k]=(x+y)%mod;
                a[i+j+k]=(x-y+mod)%mod;
            }
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;++i)scanf("%lld",a+i);
    for(int i=0;i<=m;++i)scanf("%lld",b+i);
    int len=1<<max((int)ceil(log2(n+m)),1);
    for(int i=0;i<len;++i){
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(max((int)ceil(log2(n+m)),1)-1));
    }

    NTT(a,len,1);
    NTT(b,len,1);
    for(int i=0;i<=len;++i)a[i]=a[i]*b[i]%mod;
    NTT(a,len,0);
    ll inv=quikpow(len,mod-2);
    for(int i=0;i<=n+m;++i)printf("%lld ",a[i]*inv%mod);
    return 0;
}