1001.Tetrahedron (数学 / 几何计算) 2020 Multi-University Training Contest 5

传送门

思路：

• 题意：从[1，n]随机生成直角四面体的三边，求顶点到底面高h的 1/(n^2).
• 官方题解思路1：
  
• 官方题解思路2：
  

代码实现：

#include
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 998244353;
const int  N = 6e6 + 5;

int t, n, a[N];

int qmi(int a, int k){
    int res = 1;
    while(k){
        if(k & 1) res = (ll)res * a % mod;
        k >>= 1;
        a = (ll)a * a % mod;
    }
    return res;
}

void Inint(){
    a[1] = 1;
    for(int i = 2; i < N; i ++)
        a[i] = (a[i-1] + qmi((i*i)%mod, mod-2))%mod;
}

signed main(){
    IOS;

    Inint();
    cin >> t;
    while(t --){
        cin >> n;
        cout << (a[n]*qmi(n, mod-2)%mod)*3%mod << endl;
    }

    return 0;
}