考虑容斥，钦定 \(i\) 行 \(j\) 列放同一种颜色，其余任意。\(i = 0\) 或 \(j = 0\) 的情况平凡，下面只考虑 \(i, j \ne 0\) 的情况。
答案为：
\[\sum\limits_{i = 1}^n \sum\limits_{j = 1}^n (-1)^{i + j + 1} \binom{n}{i} \binom{n}{j} 3^{(n - i)(n - j) + 1} \]\[= -\sum\limits_{i = 1}^n \binom{n}{i} (-1)^i \sum\limits_{j = 1}^n (-1)^j \binom{n}{j} 3^{n^2 + 1} \times 3^{-ni} \times 3^{-nj} \times 3^{ij} \]\[= - 3^{n^2 + 1} \sum\limits_{i = 1}^n \binom{n}{i} (-1)^i 3^{-ni} \sum\limits_{j = 1}^n (-1)^j \binom{n}{j} 3^{(-n + i)j} \]\[= - 3^{n^2 + 1} \sum\limits_{i = 1}^n \binom{n}{i} (-1)^{n + i} 3^{-ni} ((3^{i - n} - 1)^n - 1) \]然后就能算了。
code
// Problem: C. Sky Full of Stars
// Contest: Codeforces - Codeforces Round 493 (Div. 1)
// URL: https://codeforces.com/problemset/problem/997/C
// Memory Limit: 256 MB
// Time Limit: 4000 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 = 1000100;
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;
}
const ll inv3 = qpow(3, mod - 2);
ll n, c[maxn];
void solve() {
scanf("%lld", &n);
c[0] = 1;
for (int i = 1; i <= n; ++i) {
c[i] = c[i - 1] * (n - i + 1) % mod * qpow(i, mod - 2) % mod;
}
ll ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans + ((i & 1) ? 1 : mod - 1) * c[i] % mod * qpow(3, n * (n - i) + i) % mod * 2 % mod) % mod;
}
ll coef = (mod - qpow(3, n * n + 1)) % mod;
for (int i = 1; i <= n; ++i) {
ll t = (qpow(inv3, n - i) + mod - 1) % mod;
t = (qpow(t, n) + ((n & 1) ? 1 : mod - 1)) % mod;
ans = (ans + (((n + i) & 1) ? mod - 1 : 1) * coef % mod * c[i] % mod * qpow(inv3, n * i) % mod * t % mod) % mod;
}
printf("%lld\n", ans);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}