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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
#461900#8049. Equal SumsmasterhuangTL 2ms11832kbC++172.2kb2024-07-03 10:04:162024-07-03 10:04:16

Judging History

你现在查看的是最新测评结果

  • [2024-07-03 10:04:16]
  • 评测
  • 测评结果:TL
  • 用时:2ms
  • 内存:11832kb
  • [2024-07-03 10:04:16]
  • 提交

answer

#include<bits/stdc++.h>
#define LL long long
#define P pair<int,int>
#define fi first
#define se second
#define fr(x) freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);
using namespace std;
const int N=505,M=N*N,mod=998244353;
int n,m,l[N],r[N],L[N],R[N],f[N][M],g[N][M],d[N],D[N],a[N],w[M],mmax;
inline int bger(int x){return x|=x>>1,x|=x>>2,x|=x>>4,x|=x>>8,x|=x>>16,x+1;}
inline int md(int x){return x>=mod?x-mod:x;}
inline int ksm(int x,int p){int s=1;for(;p;(p&1)&&(s=1ll*s*x%mod),x=1ll*x*x%mod,p>>=1);return s;}
inline void init(int mmax)
{
	for(int i=1,j,k;i<mmax;i<<=1)
		for(w[j=i]=1,k=ksm(3,(mod-1)/(i<<1)),j++;j<(i<<1);j++)
			w[j]=1ll*w[j-1]*k%mod;
}
inline void DNT(int *a,int mmax)
{
	for(int i,j,k=mmax>>1,L,*W,*x,*y,z;k;k>>=1)
		for(L=k<<1,i=0;i<mmax;i+=L)
			for(j=0,W=w+k,x=a+i,y=x+k;j<k;j++,W++,x++,y++)
				*y=1ll*(*x+mod-(z=*y))* *W%mod,*x=md(*x+z);
}
inline void IDNT(int *a,int mmax)
{
	for(int i,j,k=1,L,*W,*x,*y,z;k<mmax;k<<=1)
		for(L=k<<1,i=0;i<mmax;i+=L)
			for(j=0,W=w+k,x=a+i,y=x+k;j<k;j++,W++,x++,y++)
				z=1ll* *W* *y%mod,*y=md(*x+mod-z),*x=md(*x+z);
	reverse(a+1,a+mmax);
	for(int inv=ksm(mmax,mod-2),i=0;i<mmax;i++) a[i]=1ll*a[i]*inv%mod;
}
inline void NTT(int *a,int *b,int n,int m)
{
	mmax=bger(n+m);DNT(a,mmax);DNT(b,mmax);
	for(int i=0;i<mmax;i++) a[i]=1ll*a[i]*b[i]%mod;IDNT(a,mmax);
}
inline void mul(int *a,int *b,int &n,int m)
{
	NTT(a,b,n,m);n+=m;
	for(int i=n+1;i<mmax;i++) a[i]=0;
}
int main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>n>>m;
	int V=0;
	for(int i=1;i<=n;i++) cin>>l[i]>>r[i],V=max(V,r[i]);
	for(int i=1;i<=m;i++) cin>>L[i]>>R[i],V=max(V,R[i]);
	f[0][0]=1;int deg=0;init(bger(n*V));
	for(int i=1;i<=n;i++)
	{
		fill(a+l[i],a+r[i]+1,1);
		memcpy(f[i],f[i-1],sizeof(f[i]));
		mul(f[i],a,deg,r[i]);
		memset(a,0,sizeof(a));d[i]=deg;
	}

	g[0][0]=1;deg=0;
	for(int i=1;i<=m;i++)
	{
		fill(a+L[i],a+R[i]+1,1);
		memcpy(g[i],g[i-1],sizeof(g[i]));
		mul(g[i],a,deg,R[i]);
		memset(a,0,sizeof(a));D[i]=deg;
	}
	for(int i=1;i<=n;i++,cout<<"\n") for(int j=1;j<=m;j++)
	{
		int ans=0;
		for(int k=0;k<=min(d[i],D[j]);k++)
		{
			ans=(ans+1ll*f[i][k]*g[j][k])%mod;
		}
		cout<<ans<<" ";
	}
	
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 11832kb

input:

2 3
1 2
2 3
1 4
2 2
1 3

output:

2 0 0 
3 4 4 

result:

ok 6 numbers

Test #2:

score: -100
Time Limit Exceeded

input:

500 500
19 458
1 480
7 485
50 461
12 476
15 461
48 466
40 453
46 467
9 458
27 478
26 472
46 459
29 490
6 500
17 487
48 484
28 472
28 459
25 480
4 491
29 481
36 460
2 491
44 499
22 473
20 458
4 483
27 471
2 496
11 461
43 450
2 478
37 466
15 459
42 482
7 451
19 455
2 453
47 475
48 450
1 474
46 471
9 4...

output:


result: