[bzoj4665]小w的喜糖_二项式反演

小w的喜糖

题目链接：https://lydsy.com/JudgeOnline/problem.php?id=4665

数据范围：略。

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题解：

二项式反演裸题。

$f_{i,j}$表示，前$i$种钦定$j$拿到自己种类糖果的方案数。

求完了之后可以二项式反演回来即可。

代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
 
const int mod = 1000000009 ;

int n, m;

ll ans;

int col[2010], s[2010], v[2010];

ll c[2010][2010], f[2010][2010], jc[2010], ine[2010], jcc[2010];

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
	int x = 0, f = 1;
	char c = nc();
	while (c < 48) {
		if (c == '-')
			f = -1;
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x * f;
}

int main() {
    n = rd();
    for (int i = 0; i <= n; i ++ ) {
        c[i][0] = 1;
        for(int j = 1; j <= i; j ++ ) {
			c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
		}
    }
    jc[0] = ine[0] = jcc[0] = jc[1] = ine[1] = jcc[1] = 1;
    for (int i = 2; i <= n; i ++ ) {
        jc[i] = (ll)jc[i - 1] * i % mod;
		ine[i] = mod - (mod / i) * ine[mod % i] % mod;
		jcc[i] = (ll)jcc[i - 1] * ine[i] % mod;
    }
    for (int i = 1; i <= n; i ++ ) {
		col[i] = rd();
	}
    sort(col + 1, col + n + 1);
    for (int i = 1; i <= n; i ++ ) {
        if (col[i] > col[i - 1]) {
			m ++ ;
		}
		v[m] ++ ;
    }
    for(int i = 1; i <= m; i ++ ) {
		s[i] = s[i - 1] + v[i];
	}
    f[0][0] = 1;
    for (int i = 1; i <= m; i ++ ) {
		for (int j = 0; j <= s[i - 1]; j ++ ) {
			for (int k = 0; k <= v[i]; k ++ ) {
        		f[i][j + k] = (f[i][j + k] + (ll)f[i-1][j] * c[v[i]][k] % mod * jc[v[i]] % mod * jcc[v[i] - k] % mod) % mod;
			}
		}
	}
    for (int i = 0; i <= n; i ++ ) {
        ans = (ans + (ll)((i & 1) ? -1 : 1) * f[m][i] * jc[n - i] + mod) % mod;
    }
    for (int i = 1; i <= m; i ++ ) {
		ans = (ll)ans * jcc[v[i]] % mod;
	}
	cout << ans << endl ;
    return 0;
}

来源：https://www.cnblogs.com/ShuraK/p/11761472.html