CF997C Sky Full of Stars

注意到一行或一列是难以下手的，实际上不满足条件的就是没有一行是相同的，没有一列是相同的。记 $G(x, y)$ 表示刚好有 $x$ 行是相同的并且刚好 $y$ 列是相同的，$F(x, y)$ 表示至少有 $x$ 列是相同的并且至少有 $y$ 列是相同的。那么有 $F(x, y) = \sum _ {i = x} ^ n \sum _ {j = y} ^ n \binom i x \binom j y G(i, j)$ ，上二项式反演，那么有 $G(x, y) = \sum _ {i = x} ^ n \sum _ {j = y} ^ n \binom i x \binom j x (-1) ^ {i + j - x - y} F(i, j)$ 。

考察如何计算 $F(x, y)$ ，如果有 $x > 0, y > 0$ ，那么存在行和列的交叉，这意味着一定所有的钦定的行列颜色是相同的，即 $3 \binom n x \binom n y 3 ^ {(n - x) (n - y)}$ ；否则，颜色可以随便选择，即 $3 ^ {x + y} \binom n x \binom n y 3 ^ {(n - x) (n - y)}$ 。

那么答案为：

$$
\begin {aligned}
& 3 ^ {n ^ 2} - G(0, 0) \\
= & 3 ^ {n ^ 2} - \sum _ {i = 0} ^ n \sum _ {j = 0} ^ n (-1) ^ {i + j} F(i, j) \\
= & 3 ^ {n ^ 2} - (\sum _ {i = 1} ^ n \sum _ {j = 1} ^ n (-1) ^ {i + j} F(i, j) + 2\sum _ {i = 1} ^ n(-1) ^ iF(i, 0) + 3 ^ {n ^ 2})\\
= & -\sum _ {i = 1} ^ n \sum _ {j = 1} ^ n (-1) ^ {i + j} F(i, j) - 2\sum _ {i = 1} ^ n(-1) ^ iF(i, 0) \\
= & -\sum _ {i = 1} ^ n \sum _ {j = 1} ^ n (-1) ^ {i + j} 3 \binom n i \binom n j 3 ^ {(n - i) (n - j)} - 2\sum _ {i = 1} ^ n(-1) ^ i 3 ^ i \binom n i 3 ^ {(n - i) n}\\
= & -3 ^ {n ^ 2 + 1} \sum _ {i = 1} ^ n (-1) ^ i \binom n i 3 ^ {-in} \sum _ {j = 1} ^ n \binom n j (-3 ^ {i - n}) ^ j - 2 \times 3 ^ {n ^ 2} \sum _ {i = 1} ^ n \binom n i (-3 ^ {1 - n}) ^ i\\
= & -3 ^ {n ^ 2 + 1} \sum _ {i = 1} ^ n (-1) ^ i \binom n i 3 ^ {-in} (\sum _ {j = 0} ^ n \binom n j (-3 ^ {i - n}) ^ j- 1) - 2 \times 3 ^ {n ^ 2} (\sum _ {i = 0} ^ n \binom n i (-3 ^ {1 - n}) ^ i - 1)\\
= & -3 ^ {n ^ 2 + 1} \sum _ {i = 1} ^ n (-1) ^ i \binom n i 3 ^ {-in} ((1 -3 ^ {i - n}) ^ n- 1) - 2 \times 3 ^ {n ^ 2} ((1 -3 ^ {1 - n}) ^ n - 1)\\
\end {aligned}
$$

查看代码

#include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e6 + 10, mod = 998244353;
int binpow (int b, LL k)
{
    k = (k % (mod - 1) + mod - 1) % (mod - 1);
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
int n;
int inv[N], fact[N], ifact[N];
void init ()
{
    inv[1] = 1;
    for (int i = 2; i < N; ++i)
        inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < N; ++i)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
    }
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
int main ()
{
    init();
    read(n);
    int res = 0;
    for (int i = 1; i <= n ;++i)
        res = (res - C(n, i) * (i & 1 ? -1ll : 1ll) * binpow(3, -(LL)i * n) % mod * (binpow(1 - binpow(3, i - n), n) - 1)) % mod;
    write((((LL)res * binpow(3, (LL)n * n + 1) - 2ll * binpow(3, (LL)n * n) * (binpow(1 - binpow(3, 1 - n), n) - 1)) % mod + mod) % mod);
    return 0;
}