cf 1342 E. Placing Rooks - 第二类斯特林数 + 组合数学

传送门
\(n\times n\)的矩阵，放置\(n\)个棋子，要满足两个条件。
首先，要满足第一个条件就必须使得每列或每行都有一个棋子。
那么只需要求出每列的情况，然后乘2就是答案。

假设每行都有一个棋子的情况下，我们讨论要几列，如果某列有\(x\)个棋子，那么就会产生贡献度为\(x - 1\)，因为每行只有一个，所以列与列不会冲突，是独立的。那么假设有\(y\)列，\(\sum_{i = 1}^y x_i - 1 = k\)，解得\(y = n - k\)，选择\(C_{n}^{n-k}\)
那么就需要\(n-k\)列。而且每列里的棋子个数都至少一个。转换下模型，把每一列的棋子个数当成球，也就是\(n\)个球放在\(n-k\)个盒子里，不允许有空盒子。因为每一列都是有排列的，存在顺序。那么球是不同的，而且每一行也是有顺序的，相当于盒子也不同，那么只能用斯特林的模型
模型是\(m!S(k,m)\)，那么答案就是\(C_{n}^{n-k} * (n-k)!S(n, n - k)\)

把斯特林用定义式\(S(n, m)=\frac{1}{m !} \sum_{i=0}^{m}(-1)^{i} * C_{m}^{i} *(m-i)^{n}\)
那么答案就是\(C_{n}^{n - k} * \sum_{i=0}^{m}(-1)^{i} * C_{m}^{i} *(m-i)^{n}\)
注意特判一下\(k = 0\)的情况，就是只下一个棋子，是\(n!\)

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define CASE int Kase = 0; cin >> Kase; for(int kase = 1; kase <= Kase; kase++)
using namespace std;
template<typename T = long long> inline T read() {
    T s = 0, f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {s = (s << 3) + (s << 1) + ch - 48; ch = getchar();} 
    return s * f;
}
#ifdef ONLINE_JUDGE
#define qaq(...) ;
#define qwq(c) ;
#else
#define qwq(a, b) for_each(a, b, [=](int x){cerr << x << " ";}), cerr << std::endl
template <typename... T> void qaq(const T &...args) {
    auto &os = std::cerr;
    (void)(int[]){(os << args << " ", 0)...};
    os << std::endl;
}
#endif
const int N = 2e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7, CM = 998244353, INF = 0x3f3f3f3f; const ll linf = 0x7f7f7f7f7f7f7f7f;
ll pow(ll a, ll b, ll p){
    ll ans = 1; a %= p;
    while(b) {
        if(b & 1) ans = ans * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return ans;
}
namespace Combination{
    const int MAXN = 2e5 + 5;
    ll fac[MAXN], invfac[MAXN], mod;
    void init(int n, ll MOD){
        fac[0] = 1; mod = MOD;
        for(int i = 1; i <= n; i++) 
            fac[i] = fac[i - 1] * i % mod;
        invfac[n] = pow(fac[n], mod - 2, mod);
        for(int i = n; i >= 1; i--)
            invfac[i - 1] = invfac[i] * i % mod;
    }
    ll C(ll n, ll m){
        if(n < 0 || m < 0) return 0;
        return n >= m ? fac[n] * invfac[n - m] % mod * invfac[m] % mod : 0;
    }
}
ll F(ll n, ll m, ll MOD){
    ll ans = 0; int base = 1;
    for(int i = 0; i <= m; i++) {
        ll now = base * Combination::C(m, m - i) % MOD * pow(m - i, n, MOD) % MOD;
        now = (now + MOD) % MOD;
        base *= -1;
        ans = (ans + now) % MOD;
    }
    return ans;
}
void solve(int kase){
    Combination::init(N - 5, CM);
    ll n = read(), k = read();
    if(k == 0) { // n!
        ll ans = 1;
        for(int i = 1; i <= n; i++) ans = ans * i % CM;
        printf("%lld\n", ans); return;
    }
    ll ans1 = Combination::C(n, n - k);
    ll ans2 = 0;
    for(int i = 0; i <= n - k; i++) {
        ll now = ((i & 1) ? -1 : 1) * Combination::C(n - k, n - k - i) % CM;
        now = (now + CM) % CM;
        now = now * pow(n - k - i, n, CM) % CM;
        ans2 = (ans2 + now) % CM;
    }
    printf("%lld\n", 2 * ans1 * ans2 % CM);
}
const bool ISFILE = 0, DUO = 0;
int main(){
    clock_t start, finish; start = clock();
    if(ISFILE) freopen("/Users/i/Desktop/practice/in.txt", "r", stdin);
    if(DUO) {CASE solve(kase);} else solve(1);
    finish = clock(); 
    qaq("\nTime:", (double)(finish - start) / CLOCKS_PER_SEC * 1000, "ms\n");
    return 0;
}