HDU-多校2-Everything Is Generated In Equal Probability(公式+逆元)

Problem Description

One day, Y_UME got an integer N and an interesting program which is shown below:



Y_UME wants to play with this program. Firstly, he randomly generates an integer n∈[1,N] in equal probability. And then he randomly generates a permutation of length n in equal probability. Afterwards, he runs the interesting program(function calculate()) with this permutation as a parameter and then gets a returning value. Please output the expectation of this value modulo 998244353.

A permutation of length n is an array of length n consisting of integers only ∈[1,n] which are pairwise different.

An inversion pair in a permutation p is a pair of indices (i,j) such that i>j and pi<pj. For example, a permutation [4,1,3,2] contains 4 inversions: (2,1),(3,1),(4,1),(4,3).

In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Note that empty subsequence is also a subsequence of original sequence.

Refer to https://en.wikipedia.org/wiki/Subsequence for better understanding.

 

 

Input

There are multiple test cases.

Each case starts with a line containing one integer N(1≤N≤3000).

It is guaranteed that the sum of Ns in all test cases is no larger than 5×104.

 

 

Output

For each test case, output one line containing an integer denoting the answer.

 

 

Sample Input

1 2 3

 

 

Sample Output

0 332748118 554580197

 

 

Source

2019 Multi-University Training Contest 2

 

 

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推导公式为(n*n-1 / 9)%998244353

套上公式求下逆元就好了

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;

ll ksm(ll x,ll y)
{
  ll ans=1;
  while(y)
  {
      if(y&1)
      {
          ans=(ans*x)%998244353;
    }
    y>>=1;
    x=(x*x)%998244353;
  }
  
  return ans;
}
int main()
{
   ll n;
   while(~scanf("%lld",&n))
   {
       
       ll ans=((n*n-1)*ksm(9,998244351))%998244353;
       printf("%lld\n",ans);
       
   }
   return 0;
}