2020 HDU Multi-University Training Contest 5

2020 HDU Multi-University Training Contest 5

1001 Tetrahedron

链接：HDU6814 Tetrahedron

题意

• 给定直角三棱锥的三边 $a,b,c\in [1,n]$ ，求到底面高 $h$ 的平方倒数的期望

思路

• $\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
• 预处理出 $E(\frac{1}{a^2})$

代码

#include 
using namespace std;
#define int long long
const int MAXN = 6e6 + 7;
const int MOD = 998244353;
// const int INF = ;
// const int DIRX[] = {};
// const int DIRY[] = {};

int ept[MAXN];

inline int getpow(int a, int b)
{
    int ret = 1;
    for (; b; b >>= 1, a = a * a % MOD)
        if (b & 1)
            ret = ret * a % MOD;
    return ret;
}

void pre()
{
    int tmp = 0;
    for (int i = 1; i <= 6e6; ++i)
    {
        tmp += getpow(i * i % MOD, MOD - 2);
        tmp %= MOD;
        ept[i] = tmp;
    }
}

int T;
int n;
int ans;

int32_t main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    pre();
    cin >> T;
    while (T--)
    {
        cin >> n;
        ans = (3 * ept[n] % MOD) * getpow(n, MOD - 2) % MOD;
        cout << ans << endl;
    }
    return 0;
}

1003 Boring Game

链接：HDU6816 Boring Game

思路

• 模拟：判断正反面、属于那张纸。

想象 $n$张纸展开的情况，每次截取上半部分，倒置(左转 $90^{\circ }$ )后将其放置于剩余部分的左侧，模拟 $k$ 次即可。

1009 Paperfolding

链接：HDU6822 Paperfolding

代码

#include
#include
using namespace std;
const long long MOD = 998244353;
long long qpow(long long n, long long p) {
    long long ret = 1;
    for (; p; p >>= 1, n = n * n % MOD)
        if (p & 1)
            ret = ret * n % MOD;
    return ret;
}
long long inv(long long a) {
    return qpow(a, MOD - 2);
}
int main()
{
    int T;
    long long n;
    cin>>T;
    while(T--){
        scanf("%lld",&n);
        long long res, tt, ni;
        res = qpow(2,n);
        tt = 2*qpow(3,n);
        ni = inv(res);
        tt = tt * ni % MOD;
        res += tt + 1;
        res %= MOD;
        printf("%lld\n",res);
    }
    return 0;
}

1012 Set1

链接：HDU6825 Set1

思路

• 组合数学：$cnt[i]=C_{i-1}^{n-i}\ast (n-i)!\ast \frac{\prod _{k=0}^{2\cdot i-n-1}{C}_{2\cdot i-n-1-k}^2}{(\frac{2\cdot i-n-1}{2})!}$
  • 第 $i$ 号后 $n-i$ 个必然是被删除的，在前 $i-1$ 个位置中选择 $n-i$ 个与后 $n-i$ 个组合，共有 $C_{i-1}^{n-i}\cdot (n-i)!$ 种组合
  • 剩余 $i-1-(n-i)=2\cdot i-n-1$ 中选 $2$ 个为 $\prod _{k=0}^{2\cdot i-n-1}{C}_{2\cdot i-n-1-k}=\frac{(2\cdot i-n-1)!}{2^{\frac{2\cdot i-n-1}{2}}}$，又对于每种两两组合其中被删除的必然按从小到大排序，故应除 $(\frac{2\cdot i-n-1}{2})!$ 重复计算的次数

代码

#include 
using namespace std;
#define int long long
const int MAXN = 5e6 + 7;;
const int MOD = 998244353;
// const int INF = ;
// const int DIRX[] = {};
// const int DIRY[] = {};

int fac[MAXN], inv[MAXN], facinv[MAXN];

void getinv(int n)
{
    fac[0] = 1;
    for (int i = 1; i < n; i++)
        fac[i] = fac[i - 1] * i % MOD;
   	// Factorial
    inv[1] = 1;
    for (int i = 2; i < n; i++)
        inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
    facinv[0] = 1;
    for (int i = 1; i < n; i++)
        facinv[i] = inv[i] * facinv[i - 1] % MOD;
    // Factorials' Muiltiplicative Inverse
}

inline int C(int n, int m)
{
    if (m > n)
        return 0;
    return fac[n] * facinv[m] % MOD * facinv[n - m] % MOD;
}
// C_n^m

inline int getpow(int a, int b)
{
    int ret = 1;
    for (; b; b >>= 1, a = a * a % MOD)
        if (b & 1)
            ret = ret * a % MOD;
    return ret;
}

int T;
int n;
int tmp;
int ans;
int cnt[MAXN];
int sum;

int32_t main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    getinv(MAXN);
    cin >> T;
    while (T--)
    {
        cin >> n;
        sum = 0;
        for (int i = 1; i <= n; ++i)
        {
            if (i <= n / 2)
                cnt[i] = 0;
            else
            {
                cnt[i] = C(i - 1, n - i) * fac[n - i] % MOD
                     * fac[2 * i - n - 1] % MOD
                     * facinv[(2 * i - n - 1) / 2] % MOD
                     * getpow(getpow(2, (2 * i - n - 1) / 2), MOD - 2) % MOD;
                sum = (sum + cnt[i]) % MOD;
            }
        }
        sum = getpow(sum, MOD - 2);
        for (int i = 1; i <= n; ++i)
        {
            ans = cnt[i] * sum % MOD;
            cout << ans << (i == n ? "\n" : " ");
        }
    }
    return 0;
}

1007 Tree

链接：HDU6820 Tree
(待补)