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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 Sums | masterhuang | TL | 2ms | 11832kb | C++17 | 2.2kb | 2024-07-03 10:04:16 | 2024-07-03 10:04:16 |
Judging History
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...