题解：P11617 递推

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15.题解：P11617 递推02-02

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前言

第一次在 OI 中见到求数列极限的题，有点意思。

但是为什么会过这么多人啊。

思路分析

做一点尝试：

\sum_{i = m} \sum_{j = 0}^{m} r_{j} a_{i - j} = 0

$\sum_{i=m} \sum_{j=0}^{m} r_ja_{i-j}=0$

然后对于相同的 $a_i$ ，合并同类项：

\sum_{i = m} \sum_{j = 0}^{m} r_{j} a_{i} + \sum_{i = 0}^{m - 1} \sum_{j = 0}^{i} r_{m - j} a_{i} = 0

$\sum_{i=m}\sum_{j=0}^{m} r_ja_i + \sum_{i=0}^{m-1}\sum_{j=0}^{i} r_{m-j}a_i=0$

然后凑出我们的目标：

\sum_{i = 0} a_{i} = \sum_{i = 0}^{m - 1} a_{i} - \frac{\sum_{i = 0}^{m - 1} \sum_{j = 0}^{i} r_{m - j} a_{i}}{\sum_{i = 0}^{m} r_{i}}

$\sum_{i=0} a_i=\sum_{i=0}^{m-1}a_i-\frac{\sum_{i=0}^{m-1}\sum_{j=0}^{i} r_{m-j}a_i}{\sum_{i=0}^{m} r_i}$

做完了，不难发现可以 $O(m^2)$ 计算和柿。

其实画一个草表也能看出来怎么做，以 $m=3$ 为例：

代码实现

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int mod=998244353;
inline int binpow(int a,int b){
	if(!b) return 1;
	int res=binpow(a,b/2);
	if(b&1) return res*res%mod*a%mod;
	else return res*res%mod;
}
int n,a[30005],r[30005],ans,sumr,suma;
signed main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
		ans=(ans+a[i])%mod;
	}
	for(int i=0;i<=n;i++){
		cin>>r[i];
		sumr=(sumr+r[i])%mod; 
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<=i;j++){
			suma=(suma+r[n-j]*a[i]%mod)%mod;
		}
	}
	ans=(ans-suma*binpow(sumr,mod-2)%mod+mod)%mod;
	cout<<ans;
	return 0;
}