CF961G Partitions-优快云博客

本文链接：https://blog.youkuaiyun.com/xmr_pursue_dreams/article/details/103502471

CF961G Partitions

题目描述

Solution

推式子：
$A n s$
$=\sum w_i\sum_{s=0}^n\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right ) \left\{ \begin{aligned} n-s \\ k-1 \end{aligned} \right\}$

把前面的 $w_i$ 先扔掉。

$=\sum_{s=0}^n\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right ) \sum_{i=0}^{k-1}\frac{(-1)^i(k-i-1)^{n-s}}{i!(k-i-1)!}s$
$=\sum_{i=0}^{k-1}\frac{(-1)^i}{i!(k-i-1)!}\sum_{s=0}^n\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right )(k-i-1)^{n-s}s$
$=\sum_{i=0}^{k-1}\frac{(-1)^i}{i!(k-i-1)!}\sum_{s=0}^n\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right )(k-i-1)^{n-s}+\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right )(k-i-1)^{n-s}(s-1)$
$=\sum_{i=0}^{k-1}\frac{(-1)^i}{i!(k-i-1)!}\sum_{s=0}^n\left ( \begin{aligned} n-1 \\ s-1 \end{aligned} \right )(k-i-1)^{n-s}+\left ( \begin{aligned} n-2 \\ s-2 \end{aligned} \right )(k-i-1)^{n-s}(n-1)$
$=\sum_{i=0}^{k-1}\frac{(-1)^i}{i!(k-i-1)!}(k-i)^{n-1}+(n-1)(k-i)^{n-2}$
$=\sum_{i=0}^{k-1}\frac{(-1)^i}{i!(k-i-1)!}+(n+k-i-1)(k-i)^{n-2}$

发现是一个卷积形式，直接 $Number\;\;Theoretic\;\;Transform$ 计算即可。

但其实上面的式子还有一个更简单的表达方法：
$Ans=\sum w_i(\left\{ \begin{aligned} n \\ k \end{aligned} \right\}+(n-1)\left\{ \begin{aligned} n-1 \\ k\;\;\; \end{aligned} \right\})$

这个式子可以用组合意义简单解释。

因此总时间复杂度为 $O (n l g n)$

#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 second

using 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;
}