bzoj 4872: [Shoi2017]分手是祝愿 [期望DP]

4872: [Shoi2017]分手是祝愿

题意：n个灯开关游戏，按i后i的约数都改变状态。随机选择一个灯，如果当前最优策略\(\le k\)直接用最优策略。问期望步数\(\cdot n! \mod 1003\)

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50% n=k 送分...从大到小选就行了...实际上送了80分...

这个期望DP没想到不应该啊

\(f[i]\)表示还有i步可以结束的期望步数

\[f[i] = \frac{i}{n} f[i-1] + \frac{n-i}{n}f[i+1] +1 \\
f[i+1] = ...
\]

但是k=0就gg了

考虑差分f，或者说\(g[i]\)表示i到i-1步的期望步数

\[g[i] = \frac{i}{n} + \frac{n-i}{n}(g[i+1] + g[i] + 1),\ g[n] = 1,\ g[i \le k]=1
\]

答案就是\(g[最优策略步数]\)啰

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 1e5+5, P = 100003, mo = P;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int n, k, a[N];
int mark[N];
ll inv[N], g[N], fac = 1;
void solve() {
	int t = 0;
	for(int i=n; i>=1; i--) {
		int p = a[i];
		for(int j=i+i; j<=n; j+=i) if(mark[j]) p ^= 1;
		if(p) mark[i] = 1, t++;
	}
	if(t <= k) {printf("%lld", t * fac %mo); return;}

	inv[1] = 1;
	for(int i=2; i<=n; i++) inv[i] = (P - P/i) * inv[P%i] %P;
	for(int i=1; i<=k; i++) g[i] = 1;
	g[n] = 1;
	for(int i=n-1; i>k; i--) g[i] = ((n-i) * g[i+1] %mo + n) * inv[i] %mo;
	ll ans = 0;
	for(int i=1; i<=t; i++) ans += g[i];
	printf("%lld", ans * fac %mo);
}

int main() {
	freopen("in", "r", stdin);
	n=read(); k=read();
	for(int i=1; i<=n; i++) a[i] = read(), fac = fac * i %mo;
	solve();
}