OSU!

OSU!

期望dp

SCUACM2022集训前训练-动态规划 - Virtual Judge (vjudge.net)

1. 若连续 x 次成功的贡献是 x，则第 \(i\) 次对答案的贡献为 \(1*p\)

2. 但连续 x 次成功的贡献是 \(x^3\), 而期望只有线性性质，对于 \(x^3\) 不可累加

3. 设第 \(i\) 个操作成功后，期望连续长度为 \(l1[i]\), 期望连续长度的平方为 \(l2[i]\)

   1. \(l1[i]=(l1[i-1]+1)*p\)
   2. \(l2[i]=(l2[i-1]+2*l1[i-1]+1)*p\)

4. 设第 \(i\) 次成功后已经连了 \(x+1\) 次，那这一次对答案的贡献就是 \((x+1)^3-x^3=3x^2+3x+1\)

   因此贡献为 \(3*l2[i]+3*l1[i]+1\)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
double dp[N], l1[N], l2[N], p;
int n;
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
	{
		scanf("%lf",&p);
		dp[i] = (3 * l2[i-1] + 3 * l1[i-1] + 1) * p;
		l1[i] = (l1[i-1] + 1) * p;
		l2[i] = (l2[i-1] + l1[i-1] * 2 + 1) * p;
	}
	double ans = 0;
	for(int i = 1; i <= n; i++) 
		ans += dp[i];
	printf("%.1f\n",ans);
	return 0;
}