题目链接:https://vjudge.net/problem/HDU-6814
题意:在[1,n]中随机取三个数a,b,c作为直角四面体的三条直角棱,求顶点d到ABC面的高的倒数平方的数学期望。
思路:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 //#include<bits/stdc++.h> 2 #include<time.h> 3 #include <set> 4 #include <map> 5 #include <stack> 6 #include <cmath> 7 #include <queue> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstring> 12 #include <utility> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 #include <list> 17 using namespace std; 18 #define eps 1e-10 19 #define PI acos(-1.0) 20 #define lowbit(x) ((x)&(-x)) 21 #define zero(x) (((x)>0?(x):-(x))<eps) 22 #define mem(s,n) memset(s,n,sizeof s); 23 #define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} 24 typedef long long ll; 25 typedef unsigned long long ull; 26 const int maxn=6e6+5; 27 const int Inf=0x7f7f7f7f; 28 const ll Mod=1e9+7; 29 const int N=3e3+5; 30 bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂 31 int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模 32 int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位,最低位编号为 0 33 int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0,否则为 -1 34 int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); } 35 ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} 36 ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} 37 int Abs(int n) { 38 return (n ^ (n >> 31)) - (n >> 31); 39 /* n>>31 取得 n 的符号,若 n 为正数,n>>31 等于 0,若 n 为负数,n>>31 等于 -1 40 若 n 为正数 n^0=n, 数不变,若 n 为负数有 n^(-1) 41 需要计算 n 和 -1 的补码,然后进行异或运算, 42 结果 n 变号并且为 n 的绝对值减 1,再减去 -1 就是绝对值 */ 43 } 44 ll binpow(ll a, ll b,ll c) { 45 ll res = 1; 46 while (b > 0) { 47 if (b & 1) res = res * a%c; 48 a = a * a%c; 49 b >>= 1; 50 } 51 return res%c; 52 } 53 void extend_gcd(ll a,ll b,ll &x,ll &y) 54 { 55 if(b==0) { 56 x=1,y=0; 57 return; 58 } 59 extend_gcd(b,a%b,x,y); 60 ll tmp=x; 61 x=y; 62 y=tmp-(a/b)*y; 63 } 64 ll mod_inverse(ll a,ll m) 65 { 66 ll x,y; 67 extend_gcd(a,m,x,y); 68 return (m+x%m)%m; 69 } 70 ll eulor(ll x) 71 { 72 ll cnt=x; 73 ll ma=sqrt(x); 74 for(int i=2;i<=ma;i++) 75 { 76 if(x%i==0) cnt=cnt/i*(i-1); 77 while(x%i==0) x/=i; 78 } 79 if(x>1) cnt=cnt/x*(x-1); 80 return cnt; 81 } 82 int mod=998244353; 83 ll a[maxn],b[maxn]; 84 void f() 85 { 86 a[1]=1; 87 b[1]=1; 88 for(int i=2;i<maxn;i++) 89 { 90 a[i]=(mod-mod/i)*a[mod%i]%mod; 91 b[i]=(b[i-1]+a[i]*a[i]%mod)%mod; 92 } 93 } 94 int main() 95 { 96 int t; 97 f(); 98 scanf("%d",&t); 99 while(t--) 100 { 101 int n; 102 scanf("%d",&n); 103 printf("%lld ",3*a[n]*b[n]%mod); 104 } 105 return 0; 106 }