题解 [CF1228E] Another Filling the Grid

传送门

直接钦定至少有多少个满足要求的话会发现剩下任意选的那些会影响一列是否合法
所以考虑钦定至少有多少个不合法的
令 \(g_{i, j}\) 为至少有 \(i\) 行 \(\neq 1\)，同时至少有 \(j\) 列 \(\neq 1\)，\(f_{i, j}\) 是恰好
有

\[f_{0, 0}=\sum\limits_{i=0}^n\sum\limits_{j=0}^n(-1)^{n+n-i-j}\binom{i}{0}\binom{j}{0}g_{i, j} \]

同时

\[g_{i, j}=\binom{n}{i}\binom{m}{j}(k-1)^{in+j(n-i)}k^{n^2-in-j(n-i)} \]

那么就可以做了

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, k;
const ll mod=1e9+7;
ll fac[N], inv[N], ans;
inline ll C(int n, int k) {return fac[n]*inv[k]%mod*inv[n-k]%mod;}
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}

ll f(int i, int j) {return C(n, i)*C(n, j)%mod*qpow(k-1, i*n+j*(n-i))%mod*qpow(k, n*n-i*n-j*(n-i))%mod;}

signed main()
{
	n=read(); k=read();
	fac[0]=fac[1]=1; inv[0]=inv[1]=1;
	for (int i=2; i<=n; ++i) fac[i]=fac[i-1]*i%mod;
	for (int i=2; i<=n; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	for (int i=2; i<=n; ++i) inv[i]=inv[i-1]*inv[i]%mod;
	for (int i=0; i<=n; ++i)
		for (int j=0; j<=n; ++j)
			ans=(ans+((n+n-i-j)&1?-1:1)*f(i, j))%mod;
	printf("%lld\n", (ans%mod+mod)%mod);

	return 0;
}