设 \(f_i\) 为 \(i\) 对情侣完全错位的方案数，那么答案为：
\[\binom{n}{k} \frac{n!}{(n - k)!} 2^k f_{n - k}
\]
分别代表选择 \(k\) 对情侣，选择它们的位置，情侣可以换位。
\(f_i\) 有递推公式：
\[f_i = 4i(i - 1) (f_{i - 1} + 2(i - 1) f_{i - 2})
\]
考虑选出两个人，另外对应的两个人的方案即可。
code
// Problem: P4931 [MtOI2018] 情侣？给我烧了！（加强版）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4931
// Memory Limit: 500 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 5000100, N = 5000000;
const ll mod = 998244353;
inline ll qpow(ll b, ll p) {
ll res = 1;
while (p) {
if (p & 1) {
res = res * b % mod;
}
b = b * b % mod;
p >>= 1;
}
return res;
}
ll n, m, fac[maxn], ifac[maxn], pw[maxn], f[maxn];
inline void init() {
pw[0] = 1;
for (int i = 1; i <= N; ++i) {
pw[i] = pw[i - 1] * 2 % mod;
}
fac[0] = 1;
for (int i = 1; i <= N; ++i) {
fac[i] = fac[i - 1] * i % mod;
}
ifac[N] = qpow(fac[N], mod - 2);
for (int i = N - 1; ~i; --i) {
ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
f[0] = 1;
for (int i = 2; i <= N; ++i) {
f[i] = 1LL * i * (i - 1) % mod * 4 % mod * (f[i - 1] + (i - 1) * 2 % mod * f[i - 2] % mod) % mod;
}
}
inline ll C(ll n, ll m) {
if (n < m || n < 0 || m < 0) {
return 0;
} else {
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
}
void solve() {
scanf("%lld%lld", &n, &m);
printf("%lld\n", C(n, m) * C(n, m) % mod * fac[m] % mod * pw[m] % mod * f[n - m] % mod);
}
int main() {
init();
int T = 1;
scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}