codeforces 235B. Let's Play Osu! （概率与期望DP）

B. Let's Play Osu!

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You're playing a game called Osu! Here's a simplified version of it. There are n clicks in a game. For each click there are two outcomes: correct or bad. Let us denote correct as "O", bad as "X", then the whole play can be encoded as a sequence of n characters "O" and "X".

Using the play sequence you can calculate the score for the play as follows: for every maximal consecutive "O"s block, add the square of its length (the number of characters "O") to the score. For example, if your play can be encoded as "OOXOOOXXOO", then there's three maximal consecutive "O"s block "OO", "OOO", "OO", so your score will be 22 + 32 + 22 = 17. If there are no correct clicks in a play then the score for the play equals to 0.

You know that the probability to click the i-th (1 ≤ i ≤ n) click correctly is pi. In other words, the i-th character in the play sequence haspi probability to be "O", 1 - pi to be "X". You task is to calculate the expected score for your play.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of clicks. The second line contains n space-separated real numbersp1, p2, ..., pn (0 ≤ pi ≤ 1).

There will be at most six digits after the decimal point in the given pi.

Output

Print a single real number — the expected score for your play. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Examples

input

3
0.5 0.5 0.5

output

2.750000000000000

input

4
0.7 0.2 0.1 0.9

output

2.489200000000000

input

5
1 1 1 1 1

output

25.000000000000000

Note

For the first example. There are 8 possible outcomes. Each has a probability of 0.125.

• "OOO"  →  32 = 9;
• "OOX"  →  22 = 4;
• "OXO"  →  12 + 12 = 2;
• "OXX"  →  12 = 1;
• "XOO"  →  22 = 4;
• "XOX"  →  12 = 1;
• "XXO"  →  12 = 1;
• "XXX"  →  0.

So the expected score is 

题目大意：每个位置pi的概率o，(1-pi)的概率是x。得分是所有o的最长连续子段的长度的平方和。

题解：概率与期望DP

同bzoj3450 维护一个最长连续长度的期望，再维护一个总分期望

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define N 100003
using namespace std;
double f[N],g[N];
int n;
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d",&n);
	f[0]=0; g[0]=0;
	for (int i=1;i<=n;i++) {
		double p; scanf("%lf",&p);
		g[i]=(g[i-1]+1)*p;
		f[i]=f[i-1]-(g[i-1]*g[i-1]);
		f[i]+=(g[i-1]+1)*(g[i-1]+1); f[i]*=p;
		f[i]+=f[i-1]*(1.0-p);
	}
	printf("%.12lf\n",f[n]);
}