bzoj5093: [Lydsy1711月賽]圖的價值

題目描述

Solution

考慮每一個點的貢獻，枚舉它的度數。
$$Ans=n?2^{\left(\genfrac{}{}{0ex}{}{n?1}{2}\right)}\sum _{i=1}^{n?1}\left(\begin{array}{r}n?1\\ i\end{array}\right)?i^k$$

現在需要解決這個部分：
${\sum }_{i=1}^n\left(\begin{array}{r}n\\ i\end{array}\right)?i^k$

看到有一個$i^k$，考慮將其展開：
$={\sum }_{i=1}^n\left(\begin{array}{r}n\\ i\end{array}\right){\sum }_j\left\{\begin{array}{r}k\\ j\end{array}\right\}\left(\begin{array}{r}i\\ j\end{array}\right)j!$
$={\sum }_{i=1}^n\frac{n!}{i!(n?i)!}{\sum }_j\left\{\begin{array}{r}k\\ j\end{array}\right\}\frac{i!}{(i?j)!}$
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}\frac{n!}{(n?i)!(i?j)!}$
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}\left(\begin{array}{r}n?j\\ n?i\end{array}\right)\frac{1}{(n?j)!}$

二項式定理：
$={\sum }_j{\sum }_i\left\{\begin{array}{r}k\\ j\end{array}\right\}{2}^{(n?j)}\frac{1}{(n?j)!}$

$NumberTheoreticTransform$即可。
時間復雜度$O(nlgn)$。

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; }

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