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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
#384384#8568. Expected DiameterzhouhuanyiWA 6090ms50672kbC++142.2kb2024-04-09 22:35:032024-04-09 22:35:03

Judging History

你现在查看的是最新测评结果

  • [2024-04-09 22:35:03]
  • 评测
  • 测评结果:WA
  • 用时:6090ms
  • 内存:50672kb
  • [2024-04-09 22:35:03]
  • 提交

answer

#include<iostream>
#include<cstdio>
#define N 2000
#define mod 998244353
using namespace std;
int read()
{
	char c=0;
	int sum=0;
	while (c<'0'||c>'9') c=getchar();
	while ('0'<=c&&c<='9') sum=sum*10+c-'0',c=getchar();
	return sum;
}
int fast_pow(int a,int b)
{
	int res=1,mul=a;
	while (b)
	{
		if (b&1) res=1ll*res*mul%mod;
		mul=1ll*mul*mul%mod,b>>=1;
	}
	return res;
}
void Adder(int &x,int d)
{
	x+=d;
	if (x>=mod) x-=mod;
	return;
}
void Adder2(int &x,int d)
{
	x+=d;
	if (x<0) x+=mod;
	return;
}
int MD(int x)
{
	return x>=mod?x-mod:x;
}
int MD2(int x)
{
	return x<0?x+mod:x;
}
int n,x,y,p,ans,fac[N+1],invfac[N+1],inv[N+1],dp[N+1][N+1],delta[N+1],W[N+1],DP[N+1][N+1],F[N+1][N+1];
int main()
{
	fac[0]=1;
	for (int i=1;i<=N;++i) fac[i]=1ll*fac[i-1]*i%mod;
	invfac[N]=fast_pow(fac[N],mod-2);
	for (int i=N-1;i>=0;--i) invfac[i]=1ll*invfac[i+1]*(i+1)%mod;
	for (int i=1;i<=N;++i) inv[i]=1ll*fac[i-1]*invfac[i]%mod;
	n=read(),x=read(),y=read(),p=1ll*x*fast_pow(y,mod-2)%mod,dp[0][1]=DP[0][1]=1;
	for (int i=1;i<=n;++i)
	{
		for (int j=1;j<=n;++j)
		{
			Adder(F[i][j],1ll*p*dp[i-1][j]%mod);
			if (i>=2) Adder(F[i][j],1ll*MD2(1-p)*dp[i-2][j]%mod);
		}
		for (int j=1;j<=n;++j) delta[j]=1ll*F[i][j]*j%mod;
		for (int j=0;j<=n;++j) W[j]=0;
		W[0]=1;
		for (int j=1;j<=n;++j)
		{
			for (int k=1;k<=j;++k) Adder(W[j],1ll*W[j-k]*delta[k]%mod);
			W[j]=1ll*W[j]*inv[j]%mod;
		}
		for (int j=1;j<=n;++j) dp[i][j]=W[j-1];
	}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			DP[i][j]=MD2(dp[i][j]-dp[i-1][j]);
	for (int i=n;i>=1;--i)
		for (int j=1;j<=n;++j)
			Adder2(F[i][j],-F[i-1][j]);
	for (int op=0;op<=1;++op)
		for (int i=0;i<=(n<<1);++i)
			for (int j=max(i-op,0);j<=min(i+op,n<<1);++j)
				for (int k=1;k<=n-1;++k)
				{
					if (!op) Adder(ans,1ll*p*DP[i][k]%mod*DP[j][n-k]%mod*inv[2]%mod*(i+j+op+1)%mod);
					else Adder(ans,1ll*MD2(1-p)*DP[i][k]%mod*DP[j][n-k]%mod*inv[2]%mod*(i+j+op+1)%mod);
				}
	for (int i=1;i<=n;++i)
	{
		Adder(ans,1ll*dp[i][n]*(i<<1)%mod);
		Adder2(ans,-1ll*dp[i-1][n]*(i<<1)%mod);
		for (int j=1;j<=n-1;++j) Adder2(ans,-1ll*dp[i-1][j]*F[i][n-j]%mod*(i<<1)%mod);
	}
	printf("%lld\n",1ll*ans*fac[n]%mod*fast_pow(n,mod-1-(n-2))%mod);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 7964kb

input:

2 1 3

output:

665496237

result:

ok 1 number(s): "665496237"

Test #2:

score: 0
Accepted
time: 1ms
memory: 7916kb

input:

3 2 3

output:

665496238

result:

ok 1 number(s): "665496238"

Test #3:

score: -100
Wrong Answer
time: 6090ms
memory: 50672kb

input:

2000 1 2

output:

720256555

result:

wrong answer 1st numbers differ - expected: '254870088', found: '720256555'