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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#494306#9141. Array Spreaducup-team3510#WA 0ms3816kbC++201.6kb2024-07-27 15:08:212024-07-27 15:08:25

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 18:58:44]
  • hack成功,自动添加数据
  • (/hack/840)
  • [2024-09-18 18:53:02]
  • hack成功,自动添加数据
  • (/hack/839)
  • [2024-07-29 03:53:23]
  • hack成功,自动添加数据
  • (/hack/753)
  • [2024-07-29 03:51:16]
  • hack成功,自动添加数据
  • (/hack/752)
  • [2024-07-29 03:50:24]
  • hack成功,自动添加数据
  • (/hack/751)
  • [2024-07-29 03:48:52]
  • hack成功,自动添加数据
  • (/hack/750)
  • [2024-07-27 15:08:25]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3816kb
  • [2024-07-27 15:08:21]
  • 提交

answer

#include <bits/stdc++.h>
#define s1 first
#define s2 second
using namespace std;
int t,n,m;
int l[2011],r[2011];
pair<int,int*> a[4011];
int mn[4011],mx[4011];
int A,B;
vector<int> v[4011];
int mxr[4011];
const int p=998244353;
inline int qpow(int bs,int ex){int ans=1;while(ex){if(ex&1)ans=1ll*ans*bs%p;ex>>=1;bs=1ll*bs*bs%p;}return ans;}
int main()
{
	scanf("%d",&t);while(t--)
	{
		scanf("%d%d",&n,&m);
		int an=0;
		for(int i=1;i<=m;++i)
		{
			scanf("%d%d",l+i,r+i);
			++an;a[an].s1=l[i],a[an].s2=l+i;
			++an;a[an].s1=r[i],a[an].s2=r+i;
		}
		sort(a+1,a+1+an);a[0].s1=-1e9;
		int k=0;
		for(int i=1;i<=an;++i)k+=a[i].s1!=a[i-1].s1,*a[i].s2=k;
		for(int i=1;i<=k+1;++i)v[i].clear(),mxr[i]=0;
		for(int i=1;i<=m;++i)v[r[i]].push_back(l[i]),mxr[l[i]]=max(mxr[l[i]],r[i]);
		int A=1,B=1;
		// for(int i=1;i<=m;++i)printf("[%d,%d] ",l[i],r[i]);putchar(10);
		for(int i=1;i<=k;++i)
		{
			for(int j=i;j<=k+1;++j)mx[j]=-1e9,mn[j]=1e9;
			mx[i-1]=0;mn[i-1]=0;
			for(int j=i;j<=k;++j)
			{
				mx[j]=max(mx[j],mx[j-1]);
				for(int l:v[j])if(l>=i)mx[j]=max(mx[j],mx[l-1]+1);
			}
			int curr=0;
			int curi=i-1;
			for(int j=1;j<=curi+1;++j)curr=max(curr,mxr[j]);
			int stp=0;
			while(curr>curi)
			{
				++stp;
				mn[curr]=min(mn[curr],stp);
				int tt=curr;
				for(int j=curi+2;j<=tt+1;++j)curr=max(curr,mxr[j]);
				curi=tt;
			}
			for(int j=n;j>=i;--j)mn[j]=min(mn[j],mn[j+1]);
			for(int j=i;j<=k;++j)if(mn[j]<1e9)
			{
				if(1ll*mx[j]*B>1ll*A*mn[j])A=mx[j],B=mn[j];
			}
		}
		printf("%lld\n",1ll*A*qpow(B,p-2)%p);
	}
	fclose(stdin);fclose(stdout);return 0;
}

Details

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Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3816kb

input:

3
3 3
1 3
2 3
1 2
12 6
2 3
5 7
1 9
4 8
1 2
7 11
4 5
3 4
2 3
1 2
4 4
1 1

output:

1
0
499122178

result:

wrong answer 2nd numbers differ - expected: '2', found: '0'