CF961G Partitions

題目描述

Solution

推式子：
$Ans$
$=\sum w_i{\sum }_{s=0}^n\left(\begin{array}{r}n?1\\ s?1\end{array}\right)\left\{\begin{array}{r}n?s\\ k?1\end{array}\right\}$

把前面的$w_i$先扔掉。

$={\sum }_{s=0}^n\left(\begin{array}{r}n?1\\ s?1\end{array}\right){\sum }_{i=0}^{k?1}\frac{(?1{)}^i(k?i?1{)}^{n?s}}{i!(k?i?1)!}s$
$={\sum }_{i=0}^{k?1}\frac{(?1{)}^i}{i!(k?i?1)!}{\sum }_{s=0}^n\left(\begin{array}{r}n?1\\ s?1\end{array}\right)(k?i?1{)}^{n?s}s$
$={\sum }_{i=0}^{k?1}\frac{(?1{)}^i}{i!(k?i?1)!}{\sum }_{s=0}^n\left(\begin{array}{r}n?1\\ s?1\end{array}\right)(k?i?1{)}^{n?s}+\left(\begin{array}{r}n?1\\ s?1\end{array}\right)(k?i?1{)}^{n?s}(s?1)$
$={\sum }_{i=0}^{k?1}\frac{(?1{)}^i}{i!(k?i?1)!}{\sum }_{s=0}^n\left(\begin{array}{r}n?1\\ s?1\end{array}\right)(k?i?1{)}^{n?s}+\left(\begin{array}{r}n?2\\ s?2\end{array}\right)(k?i?1{)}^{n?s}(n?1)$
$={\sum }_{i=0}^{k?1}\frac{(?1{)}^i}{i!(k?i?1)!}(k?i{)}^{n?1}+(n?1)(k?i{)}^{n?2}$
$={\sum }_{i=0}^{k?1}\frac{(?1{)}^i}{i!(k?i?1)!}+(n+k?i?1)(k?i{)}^{n?2}$

發現是一個卷積形式，直接$NumberTheoreticTransform$計算即可。

但其實上面的式子還有一個更簡單的表達方法：
$$Ans=\sum w_i(\left\{\begin{array}{r}n\\ k\end{array}\right\}+(n?1)\left\{\begin{array}{r}n?1\\ k\end{array}\right\})$$

這個式子可以用組合意義簡單解釋。

因此總時間復雜度為$O(nlgn)$

#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=1e9+7; 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 fac[MAXN],inv[MAXN]; 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; } int solve(int n,int m) {int ret=0;for (int i=0;i<=m;i++)ret=upd(ret,1ll*(i&1?mods-1:1)*inv[i]%mods*quick_pow(m-i,n)%mods*inv[m-i]%mods);return ret; } int main() {int n=read(),m=read(),sum=0;for (int i=1;i<=n;i++) sum=upd(sum,read());fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;inv[n]=quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mods;printf("%d\n",1ll*sum*upd(solve(n,m),1ll*(n-1)*solve(n-1,m)%mods)%mods);return 0; }

總結

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