[LOJ #6433]「PKUSC2018」最大前缀和

题目大意：给你一个$n(nleqslant20)$项的数列$A$，设重排后的数列为$A'$，令$pre_p=sumlimits_{i=1}^pA'_i$，求$max{pre_i}$的期望，乘$n!$

题解：令$f_S$为选$S$集合的数，重排后满足$max{pre_i}=sumlimits_{i=1}^{|S|}S_i$的方案数，$g_S$为选$S$集合数，重排后满足$max{pre_i}leqslant0$的方案数。发现若数列$B$满足$sumlimits_{i=1}^{|B|}B_i>0$，那么任意在它前面插入一个数，都满足$f$的条件。若数列$B$满足$max{pre_i}leqslant0$，在它后面插入一个数后，只要$sumlimits_{i=1}^{|B|}B_ileqslant0$，就行了。

答案是$sumlimits_{S}sum_Sf_Sg_{ar S}$。

卡点：无

C++ Code：

#include <cstdio>
#define maxn 20
#define N (1 << maxn)
const int mod = 998244353;
inline void reduce(int &x) { x += x >> 31 & mod; }

int n;
int s[maxn], f[N], g[N], sum[N];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf("%d", s + i);
		f[1 << i] = 1;
		g[1 << i] = s[i] <= 0;
	}
	const int U = 1 << n, I = U - 1;
	for (int i = 1; i < U; ++i) sum[i] = sum[i & i - 1] + s[__builtin_ctz(i)];
	g[0] = 1;
	for (int i = 0; i < U; ++i) if (__builtin_popcount(i) > 1) {
		for (int j = i; j; j &= j - 1) {
			int k = __builtin_ctz(j), l = i ^ 1 << k;
			if (sum[i] <= 0) reduce(g[i] += g[l] - mod);
			if (sum[l] > 0) reduce(f[i] += f[l] - mod);
		}
	}
	for (int i = 0; i < U; ++i) reduce(sum[i] %= mod);
	int ans = 0;
	for (int i = 1; i < U; ++i) reduce(ans += static_cast<long long> (sum[i]) * f[i] % mod * g[I ^ i] % mod - mod);
	printf("%d
", ans);
	return 0;
}