Blog of RuSun

OneProblemIsDifficultBecauseYouDontKnowWhyItIsDiffucult

CF997C Sky Full of Stars

LuoGu: CF997C Sky Full of Stars

CF: C. 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)=i=xnj=yn(ix)(jy)G(i,j) ,上二项式反演,那么有 G(x,y)=i=xnj=yn(ix)(jx)(1)i+jxyF(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(nx)(ny)3(nx)(ny) ;否则,颜色可以随便选择,即 3x+y(nx)(ny)3(nx)(ny)

那么答案为:

3n2G(0,0)=3n2i=0nj=0n(1)i+jF(i,j)=3n2(i=1nj=1n(1)i+jF(i,j)+2i=1n(1)iF(i,0)+3n2)=i=1nj=1n(1)i+jF(i,j)2i=1n(1)iF(i,0)=i=1nj=1n(1)i+j3(ni)(nj)3(ni)(nj)2i=1n(1)i3i(ni)3(ni)n=3n2+1i=1n(1)i(ni)3inj=1n(nj)(3in)j2×3n2i=1n(ni)(31n)i=3n2+1i=1n(1)i(ni)3in(j=0n(nj)(3in)j1)2×3n2(i=0n(ni)(31n)i1)=3n2+1i=1n(1)i(ni)3in((13in)n1)2×3n2((131n)n1)

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
#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;
}

Gitalk 加载中 ...