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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
#84808#990. Colorful ComponentsKostlinWA 6ms4584kbC++142.0kb2023-03-06 19:16:192023-03-06 19:16:56

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.
  • [2023-03-06 19:16:56]
  • 评测
  • 测评结果:WA
  • 用时:6ms
  • 内存:4584kb
  • [2023-03-06 19:16:19]
  • 提交

answer

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define fi first
#define sc second 
#define mkp make_pair
#define pii pair<int,int>
typedef long long ll;
const int N=305,mod=998244353,inv2=(mod+1)/2;
inline int read() {
	int x=0,flag=0;char ch=getchar();
	while(ch<'0'||ch>'9') {flag|=(ch=='-');ch=getchar();}
	while('0'<=ch&&ch<='9') {x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return flag?-x:x;
}
inline int mx(int x,int y) {return x>y?x:y;}
inline int mn(int x,int y) {return x<y?x:y;}
inline void swp(int&x,int&y) {x^=y^=x^=y;}
inline int as(int x) {return x>0?x:-x;}

int n,K,c[N],tr[N],ifac[N],C[N][N],f[N][N],pw[N][N];
int F[N],G[N];
inline int ksm(int a,int b=mod-2) {
	int ans=1;
	while(b) {
		if(b&1) ans=1ll*ans*a%mod;
		a=1ll*a*a%mod;
		b>>=1;
	} return ans;
}
inline void prep() {
	tr[1]=1;
	for(int i=2;i<=n;++i) tr[i]=ksm(i,i-2);
	ifac[0]=1;
	for(int i=1;i<=n;++i) ifac[i]=1ll*ifac[i-1]*ksm(i)%mod;
	C[0][0]=1;
	for(int i=1;i<=n;++i) {
		C[i][0]=C[i][i]=1;
		for(int j=1;j<i;++j)
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	for(int i=1;i<=n;++i) {
		pw[i][0]=1;
		for(int j=1;j<=n;++j)
			pw[i][j]=1ll*pw[i][j-1]*i%mod;
	}
	f[0][0]=1;
	for(int p=1;p<=n;++p)
		for(int i=1;i<=n;++i)
			for(int j=1;j<=mn(i,K);++j)
				f[p][i]=(f[p][i]+1ll*f[p-1][i-j]*C[i][j]%mod*tr[j]%mod*j)%mod;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j) 
			f[i][j]=1ll*f[i][j]*ifac[i]%mod;
}
int main() {
	n=read();K=read();
	for(int i=1,x;i<=n;++i) {
		x=read();
		++c[x];
	}
	prep();
	F[0]=1;
	int tot=0;
	for(int i=1;i<=n;++i) {
		if(!c[i]) continue;
		++tot;
		for(int j=0;j<=n;++j)
			for(int k=1;k<=c[i];++k)
				G[j+k]=(G[j+k]+1ll*F[j]*f[k][c[i]]%mod*pw[n-c[i]][k-1])%mod;
		for(int j=0;j<=n;++j) F[j]=G[j],G[j]=0;
	}
	if(tot==1) {
		if(K==n) printf("%d\n",ksm(n,n-2));
		else printf("0\n");
		return 0;
	}
	int ans=0;
	for(int i=tot;i<=n;++i) ans=(ans+1ll*F[i]*ksm(n,tot-2))%mod;
	printf("%d\n",ans);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3660kb

input:

5 3
1
1
3
1
5

output:

125

result:

ok "125"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3460kb

input:

4 2
2
1
1
1

output:

7

result:

ok "7"

Test #3:

score: -100
Wrong Answer
time: 6ms
memory: 4584kb

input:

300 16
2
2
2
4
5
1
5
3
5
4
2
1
4
4
4
5
2
1
5
4
3
4
5
3
5
5
1
3
1
1
2
5
5
3
3
2
5
2
3
2
2
4
2
2
2
4
4
2
2
4
1
3
3
4
1
3
3
4
3
4
3
5
5
4
3
3
1
2
1
2
5
2
2
4
3
3
5
3
2
4
3
5
1
4
5
5
2
3
2
3
4
4
5
5
5
5
4
5
3
2
4
4
4
3
5
3
1
1
3
5
5
4
5
2
5
5
5
2
2
2
3
1
5
4
1
4
3
5
1
4
4
2
5
2
2
4
5
3
4
3
3
4
2
5
1
1
3...

output:

840745403

result:

wrong answer 1st words differ - expected: '540253743', found: '840745403'