2020杭电多校第五场

1001.Tetrahedron

1/h² = 1/a²+1/b²+1/c²

可以算出 ans=（3∑1 / i²）/n

On预处理一下1 / i²

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

int mod = 998244353;
int inv[6000010], inv2[6000010];
int n;
inline void solve(int T)
{
    cin >> n;
    cout << 3ll * inv2[n] * inv[n] % mod << endl;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    inv[1] = inv2[1] = 1;
    rep(i, 2, 6000000)
    {
        inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
        inv2[i] = (inv2[i - 1] + (ll)inv[i] * inv[i]) % mod;
    }

    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

1009.Paperfolding

横着折x次竖着折y次的分割数为 2^x+2^y+2^x+y+1

结果的概率就表示为 （∑C(n,i)*(2ⁱ+2^n-i+2ⁿ+1)）/ 2ⁿ

化简就成了 3ⁿ / 2^n-1+2ⁿ+1

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

ll mod = 998244353;
ll ksm(ll a, ll b)
{
    ll res = 1;
    for (; b; b >>= 1, a = a * a % mod)
        if (b & 1)
            res = res * a % mod;
    return res;
}
ll n;

inline void solve(int T)
{
    cin >> n;
    if (n == 0)
        puts("4");
    else
        cout << (1ll * ksm(3, n) * ksm(499122177, n - 1) % mod + ksm(2, n) + 1) % mod << endl;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}