atcoderI - Coins ( 機率DP)

I - Coins

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100 points

Let NN be a positive odd number.

There are NN coins, numbered 1,2,…,N1,2,…,N. For each ii (1≤i≤N1≤i≤N), when Coin ii is tossed, it comes up heads with probability pipi and tails with probability 1−pi1−pi.

Taro has tossed all the NN coins. Find the probability of having more heads than tails.

NN is an odd number.

1≤N≤29991≤N≤2999

pipi is a real number and has two decimal places.

0<pi<10<pi<1

Input is given from Standard Input in the following format:

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10−910−9.

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The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Head)(Coin1,Coin2,Coin3)=(Head,Head,Head) is 0.3×0.6×0.8=0.1440.3×0.6×0.8=0.144;

The probability of having (Coin1,Coin2,Coin3)=(Tail,Head,Head)(Coin1,Coin2,Coin3)=(Tail,Head,Head) is 0.7×0.6×0.8=0.3360.7×0.6×0.8=0.336;

The probability of having (Coin1,Coin2,Coin3)=(Head,Tail,Head)(Coin1,Coin2,Coin3)=(Head,Tail,Head) is 0.3×0.4×0.8=0.0960.3×0.4×0.8=0.096;

The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Tail)(Coin1,Coin2,Coin3)=(Head,Head,Tail) is 0.3×0.6×0.2=0.0360.3×0.6×0.2=0.036.

Thus, the probability of having more heads than tails is 0.144+0.336+0.096+0.036=0.6120.144+0.336+0.096+0.036=0.612.

Outputs such as ​<code>​0.500​</code>​, ​<code>​0.500000001​</code>​ and ​<code>​0.499999999​</code>​ are also considered correct.

double p[maxn];

double dp[3050][3050];

int n;

　　

dp[1][1]=p[1];

dp[1][0]=1.0000000-p[1];

注意處理下邊界情況就好了。

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