智慧题——规律题

遇到数学的式子我们就先手玩10分钟，经验之谈

当\(n=1\)

\(\dfrac{1}{a}\)

当\(n=2\)

\(\dfrac{1}{a}\left( \dfrac{1}{a}+\dfrac{1}{b}\right) +\dfrac{1}{b}\left( \dfrac{1}{a}+\dfrac{1}{b}\right)\)

化简一下就是\(\dfrac{1}{ab}\)

当\(n=3\)

我们枚举出来他们的全排列

a b c
a c b
b a c 
b c a
c a b
c b a

\(\dfrac{1}{a}\left( \dfrac{1}{a}+\dfrac{1}{b}\right) \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{1}{a}\left( \dfrac{1}{a}+\dfrac{1}{c}\right) \left( \dfrac{1}{a}+\dfrac{1}{c}+\dfrac{1}{b}\right)+\dfrac{1}{b}+( \dfrac{1}{b}+\dfrac{1}{a})+\left( \dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{c}\right)+\dfrac{1}{b}+( \dfrac{1}{b}+\dfrac{1}{c})+\left( \dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{c}\right)\)

\(+\dfrac{1}{c}\left( \dfrac{1}{c}+\dfrac{1}{a}\right) \left( \dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c}\left( \dfrac{1}{c}+\dfrac{1}{b}\right) \left( \dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\right)\)

化简一下我们会得到\(\dfrac{1}{abc}\)

用我们的数学归纳法我们会得到推广的结论，答案就是\(n\)个数的乘积

但是由于乘积在分母上，所以我们需要用逆元这个工具将它翻到分子上

代码

#include<bits/stdc++.h>
using namespace std;
long long n;
long long x;
const int mod=998244353;
long long ans;
int qmi(int a,int b)
{
  int res=1;
  while(b)
  {
    if(b&1) ans=(long long)res*a%mod;
    a=(long long)a*a%mod;
    b>>=1;
  }
  return res;
}


int main()
{
  scanf("%lld",&n);
  scanf("%lld",&ans);
  for(int i=2;i<=n;i++)
  {
    scanf("%lld",&x);
    ans=(long long)ans*x%mod;
  }
  cout<<qmi(ans,mod-2)<<endl;
  return 0;
}