敲了一份NTT模板,在很多时候答案需要取余的时候NTT有较好的的效果.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(LL i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-10
#define pii pair<LL,LL>
LL read(){
LL x=0,f=1,ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f*x;
}
const LL maxn=802200,maxm=20000,mod=(LL)(1e9+7+0.1),inf=(LL)(1e15);
template<class T>bool cmax(T& a,T b){return a<b?a=b,true:false;}
template<class T>bool cmin(T& a,T b){return a>b?a=b,true:false;}
LL n,m;
namespace NTT{
const LL maxn=1000400;
LL r,P,H=0,L=1,R[maxn],w[maxn];
LL a[maxn],b[maxn];
LL fast(LL a,LL b){
LL ans=1;
for(;b;b>>=1,a=a*a%P)
if(b&1)ans=ans*a%P;
return ans;
}
LL prime[maxn],tail=0,B[maxn],limit=(LL)(1e6+1);
void getprime(){
up(i,2,limit){
if(!B[i])prime[++tail]=i;
for(LL j=1;j<=tail&&prime[j]*i<=limit;j++){
B[i*prime[j]]=1;
if(i%prime[j]==0)break;
}
}
}
LL q[maxn],head=0;
void getr(LL mod){
if(mod==998244353){r=3;P=mod;return;}
getprime();
P=mod;LL N=mod-1,D=P-1;
for(LL i=1;prime[i]*prime[i]<=N;i++){
if(D==1)break;
if(D%prime[i]==0)head++,q[head]=prime[i];
while(D%prime[i]==0)
D/=prime[i];
}
if(D!=1)q[++head]=D;
bool flag=0;
up(i,2,N){
flag=0;
up(j,1,head)if(fast(i,(mod-1)/q[j])==1){flag=1;break;}
if(!flag){
r=i;
break;
}
}
}
void NTT(LL* a,bool flag){
up(i,0,n)if(i<R[i])swap(a[i],a[R[i]]);
for(LL len=2;len<=L;len<<=1){
LL g=fast(r,(P-1)/len),l=len>>1;
if(flag)g=fast(g,P-2);
up(i,1,l)w[i]=w[i-1]*g%P;
for(LL st=0;st<L;st+=len){
for(LL k=0;k<l;k++){
LL x=a[st+k],y=w[k]*a[st+k+l]%P;
a[st+k]=(x+y)%P;a[st+k+l]=(x-y+P)%P;
}
}
}
if(flag){
LL inv=fast(L,P-2);
up(i,0,L-1)a[i]=a[i]*inv%P;
}
}
LL solve(LL* c,LL* d,LL n,LL m,LL* ch){
up(i,0,n-1)a[i]=c[i];
up(i,0,m-1)b[i]=d[i];
for(H=0,L=1;L<n+m-1;H++)L<<=1;
w[0]=1;
up(i,n,L)a[i]=0;
up(i,m,L)b[i]=0;
up(i,0,L)R[i]=(R[i>>1]>>1)|((i&1)<<H-1);
NTT(a,0);
NTT(b,0);
up(i,0,L)a[i]=a[i]*b[i]%P;
NTT(a,1);
up(i,0,n+m-2)ch[i+1]=a[i];
}
};
LL a[maxn],b[maxn],ch[maxn];
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
n=read();m=read();
n++,m++;
up(i,0,n-1)a[i]=read();
up(i,0,m-1)b[i]=read();
NTT::getr(998244353LL);
NTT::solve(a,b,n,m,ch);
up(i,1,n+m-1)printf("%lld ",ch[i]);
return 0;
}