【学习笔记】CF961G Partitions

不难推出式子：$$\sum a_i\sum _{i=1}^{n}i\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)\left\{\begin{array}{c}n-i\\ k-1\end{array}\right\}$$

注意到模数是$10^9+7$因此正解显然不是多项式。况且多项式对我这个萌新不太友好

奈何我是丝薄

我们要算$$\sum (i+1)\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left\{\begin{array}{c}n-i\\ k-1\end{array}\right\}$$

把$i+1$拆开

显然$$\sum i\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left\{\begin{array}{c}n-i\\ k-1\end{array}\right\}=n\left\{\begin{array}{c}n\\ k\end{array}\right\}$$

$$\sum \left(\genfrac{}{}{0ex}{}{n}{i}\right)\left\{\begin{array}{c}n-i\\ k-1\end{array}\right\}=k\left\{\begin{array}{c}n\\ k\end{array}\right\}+\left\{\begin{array}{c}n\\ k-1\end{array}\right\}$$

所以$$\sum (i+1)\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left\{\begin{array}{c}n-i\\ k-1\end{array}\right\}=(n+k)\left\{\begin{array}{c}n\\ k\end{array}\right\}+\left\{\begin{array}{c}n\\ k-1\end{array}\right\}$$

利用通项公式直接算即可。

坑点：把系数$i$抄掉了，以及没有考虑到$i=0$这一项。于是自闭了一个晚上

我真是纯纯的fw啊

#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define db long double
#define cpx complex<db>
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
const int mod=1e9+7;
int n,K;
ll a[200005],fac[200005],inv[200005],res;
ll pw(ll x,ll y=mod-2){
	ll z(1);
	for(;y;y>>=1){
		if(y&1)z=z*x%mod;
		x=x*x%mod;
	}return z;
}
void init(int n){
	fac[0]=1;for(int i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod;
	inv[n]=pw(fac[n]);for(int i=n;i>=1;i--)inv[i-1]=inv[i]*i%mod;
}
ll binom(int x,int y){
	return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
ll S(int n,int K){
	ll res(0);
	for(int i=0;i<=K;i++){
		if(K-i&1)res=(res-binom(K,i)*pw(i,n))%mod;
		else res=(res+binom(K,i)*pw(i,n))%mod;
	}return res*inv[K]%mod;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin>>n>>K,init(n);for(int i=1;i<=n;i++)cin>>a[i],res=(res+a[i])%mod;
	res=((n+K-1)*S(n-1,K)+S(n-1,K-1))%mod*res%mod;
	cout<<(res+mod)%mod;
}