提交记录 15212


用户 题目 状态 得分 用时 内存 语言 代码长度
asd_a 1002i. 【模板题】多项式乘法 Accepted 100 84.179 ms 8556 KB C++ 2.24 KB
提交时间 评测时间
2020-12-14 12:07:02 2020-12-14 12:07:06
#include<bits/stdc++.h>
using namespace std;
const int mod=998244353;
const int gen=3,ivg=332748118;
const int N=1<<21|5;
inline int fp(int x,int y){int ans=1;for(;y;y>>=1,x=1ll*x*x%mod)(y&1)&&(ans=1ll*ans*x%mod);return ans;}
namespace Poly{
	struct lsp{
		int A[1<<5|5];
		lsp(int x=0){
			A[22]=fp(x,mod>>22);
			for(int i=22;i;i--)A[i-1]=1ll*A[i]*A[i]%mod;
		}inline int operator ()(const int x)const{return A[x];}
	}pw[2]={ivg,gen};
	int rev[N];
	inline void ntt(int *a,int lim,int fl=1){
		for(int i=0;i<lim;i++){
			rev[i]=(rev[i>>1]>>1)|((i&1)?(lim>>1):0);
			(i<rev[i])&&(swap(a[i],a[rev[i]]),0);
		}for(int k=1,o=1;k<lim;k<<=1,++o){long long w=1;
			for(int i=0;i<k;i++,w=w*pw[fl](o)%mod){long long t=w*a[i+k];
				for(int j=0;j<lim;t=w*a[i+k+(j+=k<<1)])
					a[i+j+k]=(a[i+j]-t)%mod,a[i+j]=(a[i+j]+t)%mod;
			}
		}if(!fl)for(int i=0,iv=fp(lim,mod-2);i<lim;i++)a[i]=1ll*a[i]*iv%mod;
	}
	int A[N],B[N];
	#define bs basic_string<int>
	struct poly{
		bs f;
		int n;
		poly(const int x=0){
			f.clear();n=0;
			(x)&&(f+=x,++n);
		}poly(const int *a,const int x){
			f.clear();n=x;
			for(int i=0;i<n;i++)f+=a[i];
		}inline int& operator [](const int x){return f[x];}
		inline int operator [](const int x)const{return f[x];}
		inline poly operator +(const poly& a)const{
			for(int i=0;i<n;i++)A[i]=f[i];for(int i=0;i<a.n;i++)B[i]=a[i];
			int m=max(n,a.n);for(int i=0;i<m;i++)A[i]=(A[i]+B[i])%mod;
			poly b=poly(A,m);for(int i=0;i<m;i++)A[i]=B[i]=0;
			return b;
		}inline poly operator -(const poly& a)const{
			for(int i=0;i<n;i++)A[i]=f[i];for(int i=0;i<a.n;i++)B[i]=a[i];
			int m=max(n,a.n);for(int i=0;i<m;i++)A[i]=(A[i]-B[i])%mod;
			poly b=poly(A,m);for(int i=0;i<m;i++)A[i]=B[i]=0;
			return b;
		}inline poly operator *(const poly& a)const{
			for(int i=0;i<n;i++)A[i]=f[i];for(int i=0;i<a.n;i++)B[i]=a[i];
			int lim=1;while(lim<n+a.n)lim<<=1;ntt(A,lim);ntt(B,lim);
			for(int i=0;i<lim;i++)A[i]=1ll*A[i]*B[i]%mod;ntt(A,lim,0);
			poly b=poly(A,n+a.n-1);for(int i=0;i<lim;i++)A[i]=B[i]=0;
			return b;
		}
	};
	#undef bs
}
using Poly::poly;
int f[N],g[N],n,m;
poly h;
int main(){
	scanf("%d%d",&n,&m);++n;++m;
	for(int i=0;i<n;i++)scanf("%d",f+i);
	for(int i=0;i<m;i++)scanf("%d",g+i);
	h=poly(f,n)*poly(g,m);
	for(int i=0;i<h.n;i++)printf("%d ",(h[i]+mod)%mod);
	return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Subtask #1 Testcase #138.76 us56 KBAcceptedScore: 0

Subtask #1 Testcase #283.807 ms8 MB + 284 KBAcceptedScore: 0

Subtask #1 Testcase #336.976 ms3 MB + 748 KBAcceptedScore: 100

Subtask #1 Testcase #436.874 ms3 MB + 740 KBAcceptedScore: 0

Subtask #1 Testcase #541.69 us56 KBAcceptedScore: 0

Subtask #1 Testcase #640.37 us56 KBAcceptedScore: 0

Subtask #1 Testcase #740.3 us56 KBAcceptedScore: 0

Subtask #1 Testcase #877.893 ms6 MB + 1012 KBAcceptedScore: 0

Subtask #1 Testcase #977.836 ms6 MB + 892 KBAcceptedScore: 0

Subtask #1 Testcase #1071.788 ms6 MB + 216 KBAcceptedScore: 0

Subtask #1 Testcase #1184.179 ms8 MB + 364 KBAcceptedScore: 0

Subtask #1 Testcase #1282.283 ms7 MB + 244 KBAcceptedScore: 0

Subtask #1 Testcase #1339.01 us56 KBAcceptedScore: 0


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