• 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(1N3000).

    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
     
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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
    ",ans);
           
       }
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/11240725.html
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