LOJ #3300「联合省选 2020 A」组合数问题

[省选联考 2020 A 卷] 组合数问题

常用手法

\[{n\choose k} k^{\underline m}={{n-m}\choose {k-m}} n^{\underline m} \]

\[(\sum_{k=0}^n f(k)\times x^k \times {n\choose k}) \bmod p \]

\[\sum_{i=0}^m a_i k^i=\sum_{j=0}^m b_i k^{\underline i} \]

通过第二类斯特林数转下降幂。

\[x^n=\sum_{i=0}^x {x\choose i}S(x,i) i!= \sum_{i=0}^x S(n,i) x^{\underline i}\\ \begin{aligned} \sum_{i=0}^m a_i k^i&=\sum_{i=0}^m a_i \sum_{j=0}^ k S(i,j)x^{\underline j}\\ &= \sum_{j=0}^ k x^{\underline j} \sum_{i=j}^m a_i S(i,j)\\ &= \sum_{i=0}^ k x^{\underline i} \sum_{j=i}^m a_j S(j,i)\\ \end{aligned}\\ b_i=\sum_{j=i}^m a_j S(j,i) \]

\(O(m^2)\) 暴力递推，转下降幂，用上面那个方法转，然后交换求和顺序，二项式定理。

\[\begin{aligned} \sum_{k=0}^n f(k)x^k {n\choose k}&=\sum_{k=0}^n x^k {n\choose k}\sum_{i=0}^m b_i k^{\underline i}\\ &=\sum_{k=0}^n x^k \sum_{i=0}^m b_i k^{\underline i}{n\choose k}\\ &=\sum_{k=0}^n x^k \sum_{i=0}^m b_i n^{\underline i}{n-i\choose k-i}\\ &= \sum_{i=0}^m b_i n^{\underline i}\sum_{k=0}^n x^k {n-i\choose k-i}\\ &= \sum_{i=0}^m b_i n^{\underline i}\sum_{k=0}^{n-i} x^{k+i} {n-i\choose k}\\ &= \sum_{i=0}^m b_ix^i n^{\underline i}\sum_{k=0}^{n-i} x^{k} {n-i\choose k}\\ &= \sum_{i=0}^m b_ix^i n^{\underline i}{(x+1)}^{n-i} \end{aligned} \]

\(O(m)\)

/*
繁华落尽，我心中仍有花落的声音。一朵，一朵，在无人的山间，轻轻飘落。
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1010, inf=0x3f3f3f3f;
int n, x, p, m, a[N], b[N], S[N][N];
inline int read(){
	int x = 0, f = 1; char ch=getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
int power(int x, int y){
	ll ret = 1;
	while(y){
		if(y & 1) ret = 1ll * ret * x % p;
		x = 1ll * x * x % p; y >>= 1;
	}
	return ret;
}
int main(){
	//freopen("ex.in", "r", stdin);
	//freopen("ex.out", "w", stdout);
	n = read(); x = read(); p = read(); m = read();
	for(int i = 0; i <= m; i++) a[i] = read();
	S[0][0]=1 % p;
	S[1][1]=1 % p;
	for(int i = 2; i <= m; i++)
		for(int j = 1; j <= i; j++)
			S[i][j] = (1ll * j * S[i-1][j] % p + S[i-1][j-1]) % p;
	for(int i = 0; i <= m; i++)
		for(int j = i; j<= m; j++)
			b[i] = (b[i] + 1ll * S[j][i] * a[j] % p) % p;
	int sx = 1, sn = 1, ans = 0;
	x = x % p;
	for(int i = 0; i <= m; i++){
		ans = (ans + 1ll * b[i] * sx % p * sn % p * power((x + 1) % p, n - i) % p) % p;
		sx = 1ll * sx * x % p;
		sn = 1ll * sn * (n - i) % p;
	}
	printf("%d\n", ans);
	return 0;
}