每日一题：数数

题目

Solution

Code

#include <stdio.h>
typedef long long ll;
const ll mod = 998244353;
const ll M = 1e7;
ll t, n, ans1, ans2, x, p[M + 5];

ll qp(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}

int main()
{
    p[1] = 1;
    for (ll i = 2; i <= M; i++)
        p[i] = (p[i - 1] * i) % mod;
    while (~scanf("%lld", &t))
    {
        while (t--)
        {
            scanf("%lld", &n);
            ans1 = 0, ans2 = 1;
            if (n & 1)
            {
                x = (n + 1) / 2;
                ans1 = (((x * n) % mod) * ((x * n) % mod)) % mod;
            }
            else
            {
                x = n / 2;
                ans1 = (((x * (n + 1)) % mod) * ((x * (n + 1)) % mod)) % mod;
            }
            ans2 = qp(p[n], 2 * n);
            printf("%lld %lld\n", ans1, ans2);
        }
    }
    return 0;
}