CF1312 D. Count the Arrays

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给要求$[1,x]$单调增,$[x,n]$单调减。可以有$m$种值，需要有两个相同元素，求有多少种这样的数列

有两个一定是独一无二的最大值，和相同的元素。

剩余只要考虑是放在右边还是左边=$2^{n-3}$
选的方案$C(m,n-1)*(n-2)，ans=C(m,n-1)\times(n-2)\times2^{n-3}$

代码 
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const int mod = 998244353;
ll qpow(ll a, ll x, ll mo)
{
    ll res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        x >>= 1;
    }
    return res;
}
int fac[N], facinv[N];
void prepare()
{
    fac[0] = 1;
    facinv[0] = 1;
    for (int i = 1; i < N; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod;
    facinv[N - 1] = qpow(fac[N - 1], mod - 2, mod);
    for (int i = N - 2; i >= 1; i--)
        facinv[i] = 1ll * facinv[i + 1] * (i + 1) % mod;
}
ll calc(ll i, ll n)
{
    if (n <= 0)
        return 0;
    if (i > n)
        return 0;
    if (i == 0)
        return 1;
    return 1ll * fac[n] * facinv[i] % mod * facinv[n - i] % mod;
}
int n, m, res;

int main()
{
    prepare();
    cin >> n >> m;
    if (n < 3)
        cout << 0 << endl;
    //cout << 233 << endl;
    else
        cout << 1ll * calc(n - 1, m) * (n - 2) % mod * qpow(2, n - 3, mod) % mod << endl;
}