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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#14491#1423. BinjahovaWA 13ms52860kbC++202.0kb2021-10-07 11:02:062022-05-17 00:34:22

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-05-17 00:34:22]
  • 评测
  • 测评结果:WA
  • 用时:13ms
  • 内存:52860kb
  • [2021-10-07 11:02:06]
  • 提交

answer

#include <bits/stdc++.h>
#define N 2000006
using namespace std;
typedef long long ll; 
typedef vector<ll> vec;
const ll mod=998244353;
int now_limit;
int rev[N];
ll wq[20][N],fac[N],inv[N];
ll ksm(ll x,ll y){
	ll res=1; while(y){ if(y&1)res=res*x%mod; x=x*x%mod; y>>=1; }
	return res;
}
void init_NTT(int limit){
	if(limit==now_limit) return;
	now_limit=limit; int l=0; while((1<<l)<limit)l++;
	for(int i=0;i<limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
	if(wq[0][0]==0){
		while(limit<1000000) limit<<=1;
		for(int mid=1,l=0;mid<limit;mid<<=1,l++){
			wq[l][0]=1,wq[l][1]=ksm(3,(mod-1)/(mid<<1));
			for(int k=2;k<mid;k++) wq[l][k]=wq[l][k-1]*wq[l][1]%mod;
		}
	}
}
void NTT(vec &a,int limit,int flag){
	init_NTT(limit);
	while(a.size()<limit) a.push_back(0);
	for(int i=0;i<limit;i++) if(i<rev[i])swap(a[i],a[rev[i]]);
	ll x,y;
	for(int mid=1,l=0;mid<limit;mid<<=1,l++)
		for(int j=0;j<limit;j+=(mid<<1))
			for(int k=0;k<mid;k++){
				x=a[j+k],y=a[j+k+mid]*wq[l][k]%mod;
				a[j+k]=(x+y)%mod,a[j+k+mid]=(x-y+mod)%mod;
			}
	if(flag==-1){
		x=ksm(limit,mod-2);
		for(int i=0;i<limit;i++) a[i]=a[i]*x%mod;
		reverse(&a[1],&a[limit]);
	}
}
vec g,f;
int n,k,inv2;
void cdq_NTT(int l,int r){
	if(r-l<2){
		if(l>k)
		for(int i=(l-k+1)/2;i+i<l;i++) f[l]=(f[l]+f[i]*f[l-i]%mod)%mod;
		int m=min((l+k)/2,l);
		for(int i=(l+1)/2;i<=m;i++) f[l]=(f[l]+f[i]*f[l-i]%mod)%mod;
		f[l]=f[l]*inv2%mod;
		return;
	}
	int mid=(l+r+1)>>1;
	cdq_NTT(l,mid);
	vec a,b; a=vector<ll>(&f[l],&f[mid]); b=vector<ll>(&f[0],&f[min(r-l,mid)]);
	int limit=1; while(limit<r-l) limit<<=1;
	NTT(a,limit,1),NTT(b,limit,1);
	for(int i=0;i<limit;i++) a[i]=a[i]*b[i]%mod;
	NTT(a,limit,-1);
	if(l!=0) for(int i=mid;i<r;i++) f[i]=(f[i]+a[i-l]+a[i-l])%mod;
	else for(int i=mid;i<r;i++) f[i]=(f[i]+a[i-l])%mod;
	cdq_NTT(mid,r);
}
int main(){
//	freopen("test.in","r",stdin);
	cin>>n>>k;
	f.resize(n+1);
	f[1]=2; inv2=ksm(2,mod-2);
	cdq_NTT(0,n+1);
	cout<<f[n]<<'\n';
}

詳細信息

Test #1:

score: 100
Accepted
time: 3ms
memory: 50636kb

input:

2 0

output:

1

result:

ok 1 number(s): "1"

Test #2:

score: 0
Accepted
time: 13ms
memory: 52720kb

input:

3 0

output:

1

result:

ok 1 number(s): "1"

Test #3:

score: 0
Accepted
time: 10ms
memory: 50800kb

input:

3 1

output:

2

result:

ok 1 number(s): "2"

Test #4:

score: 0
Accepted
time: 3ms
memory: 51568kb

input:

4 0

output:

2

result:

ok 1 number(s): "2"

Test #5:

score: 0
Accepted
time: 2ms
memory: 50772kb

input:

4 1

output:

3

result:

ok 1 number(s): "3"

Test #6:

score: 0
Accepted
time: 0ms
memory: 52860kb

input:

4 2

output:

5

result:

ok 1 number(s): "5"

Test #7:

score: 0
Accepted
time: 7ms
memory: 52692kb

input:

6 2

output:

23

result:

ok 1 number(s): "23"

Test #8:

score: -100
Wrong Answer
time: 8ms
memory: 51536kb

input:

7 42

output:

280756269

result:

wrong answer 1st numbers differ - expected: '132', found: '280756269'