QOJ.ac

QOJ

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
#88561#2573. Two Permutationss8194272WA 10ms12504kbC++141.9kb2023-03-16 15:34:572023-03-16 15:35:00

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.
  • [2023-03-16 15:35:00]
  • 评测
  • 测评结果:WA
  • 用时:10ms
  • 内存:12504kb
  • [2023-03-16 15:34:57]
  • 提交

answer

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<bitset>
#include<random>
#include<cmath>
#include<ctime>
#include<queue>
#include<map>
#include<set>

#define fi first
#define se second
#define max Max
#define min Min
#define abs Abs
#define lc (x<<1)
#define rc (x<<1|1)
#define mid ((l+r)>>1)
#define pb(x) push_back(x)
#define lowbit(x) ((x)&(-(x)))
#define fan(x) ((((x)-1)^1)+1)
#define mp(x,y) make_pair(x,y)
#define clr(f,n) memset(f,0,sizeof(int)*(n))
#define cpy(f,g,n) memcpy(f,g,sizeof(int)*(n))
#define SZ(x) ((int)(x.size()))
#define INF 0x3f3f3f3f

using namespace std;

inline int read()
{
	int ans=0,f=1;
	char c=getchar();
	while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){ans=(ans<<1)+(ans<<3)+c-'0';c=getchar();}
	return ans*f;
}

inline void write(int x)
{
	if(x<0) putchar('-'),x=-x;
	if(x/10) write(x/10);
	putchar((char)(x%10)+'0');
}

template<typename T>inline T Abs(T a){return a>0?a:-a;};
template<typename T,typename TT>inline T Min(T a,TT b){return a<b?a:b;}
template<typename T,typename TT> inline T Max(T a,TT b){return a<b?b:a;}

const int N=105,mod=998244353;
int n,k,f[2][N][N*N];

signed main()
{
	n=read();k=read();
	int now=0;
	f[0][0][0]=1;
	for(int i=0;i<n;++i)
	{
		memset(f[now^1],0,sizeof(f[now^1]));
		for(int j=0;j<=i;++j)
			for(int p=0;p<=k;++p)
				if(f[now][j][p])
				{
					if(j+1<=i) f[now^1][j+2][p]=(f[now^1][j+2][p]+f[now][j][p]*1ll*(i-j)%mod*(i-j))%mod;
					if(p+i+1<=k) f[now^1][j+1][p+n-i]=(f[now^1][j+1][p+n-i]+f[now][j][p]*1ll*(i-j)%mod*(n-i)*2)%mod;
					if(p+(i+1)*2<=k&&n-i-1>=0) f[now^1][j][p+(n-i)*2]=(f[now^1][j][p+(n-i)*2]+f[now][j][p]*1ll*(n-i)%mod*(n-i-1))%mod;
					if(n-i*2+j>=0) f[now^1][j+1][p+n-i]=(f[now^1][j+1][p+n-i]+f[now][j][p]*1ll*(n-i*2+j))%mod;
				}
		now^=1;
	}
	write(f[now][n][k]);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 2ms
memory: 12240kb

input:

2 4

output:

2

result:

ok 1 number(s): "2"

Test #2:

score: 0
Accepted
time: 10ms
memory: 12504kb

input:

3 7

output:

12

result:

ok 1 number(s): "12"

Test #3:

score: 0
Accepted
time: 2ms
memory: 12496kb

input:

4 10

output:

24

result:

ok 1 number(s): "24"

Test #4:

score: -100
Wrong Answer
time: 8ms
memory: 12320kb

input:

4 14

output:

288

result:

wrong answer 1st numbers differ - expected: '96', found: '288'