[Luogu4921] lovers? Burn It for me! [Dislocation arrangement]

Test instructions

Test instructions very clear \ Funny

Analysis

• For each query \ (k\) , remember \ ( g (x) \) indicates that \ (x\) couples are staggered by the total number of scenarios, then the answer can be written as follows:
  \[{ans}_k= \binom{n}{k}\times a_n^k\times 2^k\times g (n-k) \]

• Consider how to ask for \ (g (x) \) (a misaligned arrangement).

• Consider the first row, a total of three cases: two men and two women or a man and a woman (not paired).

  • Two male : the number of options for selecting two males in sequence is \ (x (x-1) \) , then consider the pairing of their spouse in the following situation:

    • If the force is not paired, then treat them as a couple to ensure that after the process is not paired (gay gay qi ), ie (g (x-1) \) .

    • If you force pairing, then select a row in the remaining \ (x-1\) row, and the two can be exchanged in order and transferred to \ (2 (x-1) \times g (x-2) \).

  • Two women : the number of programmes is clearly the same as for the two men.

  • a man and a woman: enumeration of a man and a woman, can be exchanged in the order of the number of programs (x (x-1) \) , the transfer is actually the same,

• So we get:\ (g (x) =4x (x-1) \times[g (x-1) +2 (x-1) \times g (x-2)]\) .

• Single processing \ (g\) complexity \ ( o (n) \) , each answer enumeration \ (k\) complexity \ (o (n) \) , the total time complexity is \ (o (n) \) .

Code

#include <bits/stdc++.h>using namespace std; #define REP (i,a,b) for (int i=a;i<=b;++i) #define PB Push_    Backtypedef long Long ll;inline int gi () {int X=0,f=1;char ch=getchar ();    while (!isdigit (CH)) {if (ch== '-') F=-1;ch=getchar ();}    while (IsDigit (ch)) {x= (x<<3) + (x<<1) +ch-48;ch=getchar ();} return x*f;} Template<typename t>inline bool Max (T &a,t b) {return a<b?a=b,1:0;} Template<typename t>inline bool Min (T &a,t b) {return b<a?a=b,1:0;} const int N=2004,mod=998244353;int t,n; ll fac[n],inv[n],invfac[n],bin[n],g[n];void Add (ll &a,ll b) {a+=b;if (a>=mod) A-=mod;}    ll Pow (ll A,ll b) {ll res=1ll;    for (; b;b>>+1,a=a*a%mod) if (b&1) Res=res*a%mod; return res;} LL C (int n,int m) {return fac[n]*invfac[n-m]%mod*invfac[m]%mod;}    int main () {fac[0]=invfac[0]=inv[1]=bin[0]=1;        Rep (i,1,n-1) {if (i^1) inv[i]= (mod-mod/i) *inv[mod%i]%mod;        Fac[i]=fac[i-1]*i%mod;        Invfac[i]=invfac[i-1]*inv[i]%mod; bin[i]=bin[i-1]*2%mod;    } g[0]=1,g[1]=0;        Rep (n,2,1000) g[n]=4ll*n* (n-1)%mod* (g[n-1]+2* (n-1) *g[n-2])%mod;    T=gi ();        while (t--) {n=gi ();    Rep (k,0,n) printf ("%lld\n", C (n,k) *c (n,k)%mod*fac[k]%mod*bin[k]%mod*g[n-k]%mod); } return 0;}

[Luogu4921] lovers? Burn It for me! [Dislocation arrangement]