提交记录 17122


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 50.637 ms 7712 KB C++11 2.83 KB
提交时间 评测时间
2021-12-01 15:33:35 2021-12-01 15:33:39
#include<bits/stdc++.h>
using namespace std;
const int mod=998244353,gen=3;
namespace math{
#ifdef DEBUG
#ifndef need_butterfly
	#define need_butterfly
#endif
#endif
	// fast power , (x and ans) must in [-mod,mod], y must in [-mod+1,mod-1] 
	typedef int i32;
	typedef unsigned int u32;
	typedef long long i64;
	typedef unsigned long long u64;
	template<typename T=int>
	inline T fsp(i64 x,int y,i64 ans=1){
		for(y<0?y+=mod-1:0;y;y>>=1,x=x*x%mod)
			y&1?ans=ans*x%mod:0;
		return ans;
	}
	inline int lg(int x){return x==0?-1:__lg(x);}
	namespace fast_number_theory_transform{
		const int maxbit=22;
		const u32 modm2=mod+mod;
		template<class T>
		inline void butterfly(T* p,int bit) {
			for(u32 i=0,j=0;i<(1u<<bit);i++){
				if(i>j)swap(p[i],p[j]);
				for(u32 l=1u<<(bit-1);(j^=l)<l;l>>=1);
			}
		}
		u32 *_p0[maxbit+1],*_p1[maxbit+1];
		inline void prep(int bit){
			static int k=0;
			for(;k<bit;k++){
				u32 *p,*q,nl=1<<k;
				u64 g=fsp(3,mod>>(k+1));
				p=_p0[k]=new u32[nl<<1];q=p+nl;
				for(int i=p[0]=1;i<nl;i++)p[i]=p[i-1]*g%mod;
				butterfly(p,k);
				for(int i=0;i<nl;i++)q[i]=(u64(p[i])<<32)/mod;
				g=fsp(g,-1);
				p=_p1[k]=new u32[nl<<1];q=p+nl;
				for(int i=p[0]=1;i<nl;i++)p[i]=p[i-1]*g%mod;
				butterfly(p,k);
				for(int i=0;i<nl;i++)q[i]=(u64(p[i])<<32)/mod;
			}
		}inline u32 ntt_mul(u32 x,u64 p,u64 q){return x*p-(q*x>>32)*mod;}
		void ntt(u32* a,int bit){
			prep(bit);
			for(int k=bit;k-->0;){
				u32 *_p=_p0[bit-k-1],*_q=_p+(1<<(bit-k-1));
				u32 *_a0=a,*_a1=a+(1<<k);
				for(int i=0;i<1<<(bit-k-1);i++,_a0+=1<<(k+1),_a1+=1<<(k+1))
					for(int j=0;j<(1<<k);++j){
						u32 x=_a0[j]-(_a0[j]>=modm2)*modm2,y=ntt_mul(_a1[j],_p[i],_q[i]);
						_a0[j]=x+y;_a1[j]=x+modm2-y;
					}
			}for(int i=0;i<(1<<bit);i++){
				a[i]-=(a[i]>=modm2)*modm2;
				a[i]-=(a[i]>=mod)*mod;
			}
		#ifdef need_butterfly
			butterfly(a,bit);
		#endif
		}
		
		void intt(u32* a,int bit){
			prep(bit);
		#ifdef need_butterfly
			butterfly(a,bit);
		#endif
			for(int k=0;k<bit;k++){
				u32 *_p=_p1[bit-k-1],*_q=_p+(1<<(bit-k-1));
				u32 *_a0=a,*_a1=a+(1<<k);
				for(int i=0;i<1<<(bit-k-1);i++,_a0+=1<<(k+1),_a1+=1<<(k+1))
					for(int j=0;j<(1<<k);++j){
						u32 x=_a0[j],y=_a1[j];
						_a0[j]=x+y-(x+y>=modm2)*modm2;
						_a1[j]=ntt_mul(x+modm2-y,_p[i],_q[i]);
					}
			}
			u64 iv=mod;iv<<=bit;iv=(iv-mod+1)>>bit;
			for(int i=0;i<(1<<bit);i++)
				a[i]=a[i]*iv%mod;
		}
	}using fast_number_theory_transform::ntt;
	using fast_number_theory_transform::intt;
}
// using math::fsp;
using math::ntt;
using math::intt;
int n,m;
unsigned a[262144],b[262144];
int main(){
// #ifndef DEBUG
// 	ios::sync_with_stdio(0);
// 	cin.tie(0);
// #endif
	cin>>n>>m;
	for(int i=0;i<=n;i++)cin>>a[i];
	for(int i=0;i<=m;i++)cin>>b[i];
	int bit=__lg(n+m)+1;
	ntt(a,bit);ntt(b,bit);
	for(int i=0;i<(1<<bit);i++)a[i]=1ll*a[i]*b[i]%mod;
	intt(a,bit);
	for(int i=0;i<=n+m;i++)cout<<a[i]<<' ';
}

CompilationN/AN/ACompile OKScore: N/A

Subtask #1 Testcase #139.25 us44 KBAcceptedScore: 0

Subtask #1 Testcase #250.637 ms7 MB + 464 KBAcceptedScore: 0

Subtask #1 Testcase #323.453 ms3 MB + 324 KBAcceptedScore: 100

Subtask #1 Testcase #423.206 ms3 MB + 312 KBAcceptedScore: 0

Subtask #1 Testcase #542.89 us44 KBAcceptedScore: 0

Subtask #1 Testcase #639.37 us44 KBAcceptedScore: 0

Subtask #1 Testcase #740.91 us44 KBAcceptedScore: 0

Subtask #1 Testcase #844.985 ms7 MB + 196 KBAcceptedScore: 0

Subtask #1 Testcase #944.97 ms7 MB + 196 KBAcceptedScore: 0

Subtask #1 Testcase #1039.301 ms6 MB + 952 KBAcceptedScore: 0

Subtask #1 Testcase #1150.416 ms7 MB + 544 KBAcceptedScore: 0

Subtask #1 Testcase #1246.916 ms6 MB + 424 KBAcceptedScore: 0

Subtask #1 Testcase #1339.89 us44 KBAcceptedScore: 0


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