6353. [simulation] to NOIP2019 (ca)

Title Description

answer

Worm combined

Since a few days ago was taught ♂ sterile, so vigorously looking out of the law

First m-1, set f [i] [j] denotes m≤i, the answer of leaf nodes j

Transfer Obviously, alsoObviously O (n ^ 3) of

---

After playing out a long way to f:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192
1 1 2 5 13 34 89 233 610 1597 4181 10946 28657 75025 196418
1 1 2 5 14 41 122 365 1094 3281 9842 29525 88574 265721 797162
1 1 2 5 14 42 131 417 1341 4334 14041 45542 147798 479779 1557649
1 1 2 5 14 42 132 428 1416 4744 16016 54320 184736 629280 2145600
1 1 2 5 14 42 132 429 1429 4846 16645 57686 201158 704420 2473785
1 1 2 5 14 42 132 429 1430 4861 16778 58598 206516 732825 2613834
1 1 2 5 14 42 132 429 1430 4862 16795 58766 207783 740924 2660139
1 1 2 5 14 42 132 429 1430 4862 16796 58785 207990 742626 2671892
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208011 742876 2674117
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742899 2674414
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674439
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

Obviously i-th row before the i + 1 is the number of normal Cattleya, item start from the i + 2 is different

it can be discovered:

\(f[1][i]=f[1][i-1]*1\)

\(f[2][i]=f[2][i-1]*2\)

\(f[3][i]=f[3][i-1]*3-f[3][i-2]*1\)

\(f[4][i]=f[4][i-1]*4-f[4][i-2]*6\)

\(f[5][i]=f[5][i-1]*5-f[5][i-2]*10+f[5][i-3]*1\)

guess:

\(f[i][j]=\sum{f[i][j-k]*a[i][k]}\)

byViolence EnumerationAfter the first few can get out of a:

1：1

2：2

3：3 -1

4：4 -3

5：5 -6 1

6：6 -10 4

7：7 -15 10 -1

8：8 -21 20 -5

Imperial \ (a [1 ... m] [0] = 1, a [1] [1] = 1, a [2] [1] = 2 \) then, can be found a magic formula:

\(a[i][j]=a[i-1][j]+a[i-2][j-1]\)

Release may then be \ (A [m] \) , then Release \ (f [m] \) to

code

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 998244353
using namespace std;

long long a[5001][5001];
long long f[5001];
int n,m,i,j,k,l;

int main()
{
    freopen("ca.in","r",stdin);
    freopen("ca.out","w",stdout);
//  freopen("a.out","w",stdout);
    
    fo(i,1,5000)
    a[i][0]=1;
    a[1][1]=1;
    a[2][1]=2;
    fo(i,3,5000)
    {
        fo(j,1,5000)
        a[i][j]=(a[i-1][j]+a[i-2][j-1])%mod;
    }
    
    scanf("%d%d",&m,&n);
    --m;
    f[1]=1;
    fo(i,2,m+1)
    {
        fo(j,1,i-1)
        f[i]=(f[i]+f[j]*f[i-j]%mod)%mod;
    }
    
    fo(i,m+2,n)
    {
        k=1;
        fd(j,i-1,1)
        {
            f[i]=(f[i]+f[j]*a[m][i-j]*k%mod)%mod;
            k=-k;
        }
    }
    
    fo(i,1,n)
    printf("%lld\n",(f[i]+mod)%mod);
    
    fclose(stdin);
    fclose(stdout);
    
    return 0;
}

Official explanations

In fact, it is not difficult, let \ (f [i] [j] \) represents i put points to the current point needs to step away from the root to the left j

So \ (f [i] [j ] \) can go to his left son, or son that went to the right point to the left of the last to go, that \ (f [i] [j ] -> f [i + 1] [j + 1], f [i + 1] [j-1] \)

Such is the nature of the transfer order dfs go by, the final answer is \ (f [2K-1] [0] \) (the last point must only go to the right)

code

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 998244353
#define min(a,b) (a<b?a:b)
using namespace std;

int f[10001][5002];
int n,m,i,j,k,l;

int main()
{
    freopen("ca.in","r",stdin);
    freopen("ca.out","w",stdout);
    
    scanf("%d%d",&m,&n);--m;
    f[1][0]=1;
    
    fo(i,1,n+n-2)
    {
        fd(j,min(m,i-1),0)
        if (f[i][j])
        {
            f[i+1][j+1]=(f[i+1][j+1]+f[i][j])%mod;
            if (j>0)
            f[i+1][j-1]=(f[i+1][j-1]+f[i][j])%mod;
        }
    }
    
    for (i=1; i<=n+n-1; i+=2)
    printf("%d\n",f[i][0]);
    
    fclose(stdin);
    fclose(stdout);
    
    return 0;
}