bzoj5093: [Lydsy1711月赛]图的价值

题目描述

Solution

考虑每一个点的贡献，枚举它的度数。
$$Ans=n\ast 2^{\left(\genfrac{}{}{0ex}{}{n-1}{2}\right)}\sum _{i=1}^{n-1}\left(\begin{array}{r}n-1\\ i\end{array}\right)\ast i^k$$

现在需要解决这个部分：
${\sum }_{i=1}^n\left(\begin{array}{r}n\\ i\end{array}\right)\ast i^k$

看到有一个$i^k$，考虑将其展开：
$={\sum }_{i=1}^n\left(\begin{array}{r}n\\ i\end{array}\right){\sum }_j\left\{\begin{array}{r}k\\ j\end{array}\right\}\left(\begin{array}{r}i\\ j\end{array}\right)j!$
$={\sum }_{i=1}^n\frac{n!}{i!(n-i)!}{\sum }_j\left\{\begin{array}{r}k\\ j\end{array}\right\}\frac{i!}{(i-j)!}$
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}\frac{n!}{(n-i)!(i-j)!}$
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}\left(\begin{array}{r}n-j\\ n-i\end{array}\right)\frac{1}{(n-j)!}$

二项式定理：
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}{2}^{(n-j)}\frac{1}{(n-j)!}$

$NumberTheoreticTransform$即可。
时间复杂度$O(nlgn)$。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int G=3;
const int Gi=(mods+1)/G;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int f[MAXN],g[MAXN],rev[MAXN],fac[MAXN],s[MAXN],Limit,L;
int quick_pow(int x,int y)
{int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;
}
int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
void Number_Theoretic_Transform(int *A,int type)
{for (int i=0;i<Limit;i++) if (i<rev[i]) swap(A[i],A[rev[i]]);for (int mid=1;mid<Limit;mid<<=1){int Wn=quick_pow(type==1?G:Gi,(mods-1)/(mid<<1));for (int j=0;j<Limit;j+=(mid<<1))for (int k=j,w=1;k<j+mid;w=1ll*w*Wn%mods,k++){int x=A[k],y=1ll*w*A[k+mid]%mods;A[k]=upd(x,y),A[k+mid]=upd(x,mods-y);}}
}
void Init(int n)
{fac[0]=1; for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;f[n]=quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) f[i]=1ll*f[i+1]*(i+1)%mods;for (int i=0;i<=n;i++){g[i]=(i&1)?mods-f[i]:f[i];f[i]=1ll*f[i]*quick_pow(i,n)%mods;}
}
int main()
{int n=read(),k=read();Init(k); Limit=1,L=0;while (Limit<=k<<1) Limit<<=1,L++; for (int i=1;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));Number_Theoretic_Transform(f,1);Number_Theoretic_Transform(g,1);for (int i=0;i<=Limit;i++) f[i]=1ll*f[i]*g[i]%mods;Number_Theoretic_Transform(f,-1);int invLimit=quick_pow(Limit,mods-2);for (int i=0;i<=k;i++) f[i]=1ll*f[i]*invLimit%mods;int ans=0;s[0]=1;for (int i=1;i<=k;i++) s[i]=1ll*s[i-1]*(n-i)%mods;for (int i=0;i<=k;i++) ans=upd(ans,1ll*f[i]*s[i]%mods*quick_pow(2,n-i-1)%mods);
//  cout<<ans<<endl;printf("%d\n",1ll*ans*n%mods*quick_pow(2,(1ll*(n-1)*(n-2)/2)%(mods-1))%mods);return 0;
}