传送门
题目
题意翻译
给出nn 个物品, 每个物品有一个权值w_iw
i
定义一个集合SS 的权值W(S)=|S|\sum\limits_{x\in S}w_xW(S)=∣S∣
x∈S
∑
w
x
定义一个划分的权值为W’®=\sum\limits_{S\subseteq R}W(S)W
′
®=
S⊆R
∑
W(S)
求将nn 个物品划分成kk 个集合的所有方案的权值和
n,k\le2\times105,w_i\le109n,k≤2×10
5
,w
i
≤10
9
感谢@Kelin 提供的翻译
题目描述
You are given a set of n n elements indexed from 1 1 to n n . The weight of i i -th element is w_{i} w
i
. The weight of some subset of a given set is denoted as . The weight of some partition R R of a given set into k k subsets is (recall that a partition of a given set is a set of its subsets such that every element of the given set belongs to exactly one subset in partition).
Calculate the sum of weights of all partitions of a given set into exactly k k non-empty subsets, and print it modulo 10^{9}+7 10
9
+7 . Two partitions are considered different iff there exist two elements x x and y y such that they belong to the same set in one of the partitions, and to different sets in another partition.
输入格式
The first line contains two integers n n and k k ( 1<=k<=n<=2·10^{5} 1<=k<=n<=2⋅10
5
) — the number of elements and the number of subsets in each partition, respectively.
The second line contains n n integers w_{i} w
i
( 1<=w_{i}<=10^{9} 1<=w
i
<=10
9
)— weights of elements of the set.
输出格式
Print one integer — the sum of weights of all partitions of a given set into k k non-empty subsets, taken modulo 10^{9}+7 10
9
+7 .
输入输出样例
输入 #1 复制
4 2
2 3 2 3
输出 #1 复制
160
输入 #2 复制
5 2
1 2 3 4 5
输出 #2 复制
645
说明/提示
Possible partitions in the first sample:
{{1,2,3},{4}} 1,2,3,4 , W®=3·(w_{1}+w_{2}+w_{3})+1·w_{4}=24 W®=3⋅(w
1
+w
2
+w
3
)+1⋅w
4
=24 ;
{{1,2,4},{3}} 1,2,4,3 , W®=26 W®=26 ;
{{1,3,4},{2}} 1,3,4,2 , W®=24 W®=24 ;
{{1,2},{3,4}} 1,2,3,4 , W®=2·(w_{1}+w_{2})+2·(w_{3}+w_{4})=20 W®=2⋅(w
1
+w
2
)+2⋅(w
3
+w
4
)=20 ;
{{1,3},{2,4}} 1,3,2,4 , W®=20 W®=20 ;
{{1,4},{2,3}} 1,4,2,3 , W®=20 W®=20 ;
{{1},{2,3,4}} 1,2,3,4 , W®=26 W®=26 ;
Possible partitions in the second sample:
{{1,2,3,4},{5}} 1,2,3,4,5 , W®=45 W®=45 ;
{{1,2,3,5},{4}} 1,2,3,5,4 , W®=48 W®=48 ;
{{1,2,4,5},{3}} 1,2,4,5,3 , W®=51 W®=51 ;
{{1,3,4,5},{2}} 1,3,4,5,2 , W®=54 W®=54 ;
{{2,3,4,5},{1}} 2,3,4,5,1 , W®=57 W®=57 ;
{{1,2,3},{4,5}} 1,2,3,4,5 , W®=36 W®=36 ;
{{1,2,4},{3,5}} 1,2,4,3,5 , W®=37 W®=37 ;
{{1,2,5},{3,4}} 1,2,5,3,4 , W®=38 W®=38 ;
{{1,3,4},{2,5}} 1,3,4,2,5 , W®=38 W®=38 ;
{{1,3,5},{2,4}} 1,3,5,2,4 , W®=39 W®=39 ;
{{1,4,5},{2,3}} 1,4,5,2,3 , W®=40 W®=40 ;
{{2,3,4},{1,5}} 2,3,4,1,5 , W®=39 W®=39 ;
{{2,3,5},{1,4}} 2,3,5,1,4 , W®=40 W®=40 ;
{{2,4,5},{1,3}} 2,4,5,1,3 , W®=41 W®=41 ;
{{3,4,5},{1,2}} 3,4,5,1,2 , W®=42 W®=42 .
思路
首先一个很显然的性质，每个数做出贡献的系数是相同的（因为是所有情况的总和）
注意到集合权值的系数为 |S|，可以看成集合中的每一个元 素都会对 wi 做一次贡献，考虑计算每个数对 wi 的贡献。 显然 i 对 wi 的贡献为{n k}，满足 j̸= i 的 j 对 wi 的贡献就是 i 和 j 被划分到同一个集合的方案数，也就是{n−1 k }所以每个 wi 的系数就是{n k}+ (n−1){n−1 k }，通过通项O (n) 算第二类斯特林数即可。 时间复杂度 O(n)
代码
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int mod=1e9+7,N=2e5+77;
ll fac[N],unfac[N],sum,ans;
ll power(ll x,ll t)
{
ll b=1;
while(t)
{
if(t&1) b=b*x%mod;
t>>=1; x=x*x%mod;
}
return b;
}
ll getS(int n,int k)
{
ll res=0;
for(int i=0; i<=k; i++)
{
ll opt=i&1?-1ll:1ll;
res=(res+opt*unfac[i]%mod*power(k-i,n)%mod*unfac[k-i]%mod+mod)%mod;
}
return res;
}
void init(int n)
{
fac[1]=1;
for(int i=2; i<=n; i++) fac[i]=fac[i-1]*i%mod;
unfac[n]=power(fac[n],mod-2);
for(int i=n-1; i>=0; i--) unfac[i]=unfac[i+1]*(i+1)%mod;
}
int main()
{
int n,k;
scanf("%d%d",&n,&k);
for(int i=1; i<=n; i++)
{
ll x;
scanf("%lld",&x);
sum=(sum+x)%mod;
}
init(n);
ll ans=(getS(n,k)+(n-1)*getS(n-1,k)%mod)%mod;
ans=ans*sum%mod;
printf("%lld",ans);
}
