atcoderI - Coins ( 概率DP)

I - Coins

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100 points

Problem Statement

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.

Constraints

• 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

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Input

Input is given from Standard Input in the following format:

NN
p1p1 p2p2 …… pNpN

Output

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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Sample Input 1 Copy

Copy

3
0.30 0.60 0.80

Sample Output 1 Copy

Copy

0.612

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.

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Sample Input 2 Copy

Copy

1
0.50

Sample Output 2 Copy

Copy

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

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Sample Input 3 Copy

Copy

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3 Copy

Copy

0.3821815872

double p[maxn];
double dp[3050][3050];
int n;

题意：给N个硬币，每一个硬币扔向空中落地是正面朝上的概率是p[i] ，让求扔了N个硬币，正面的数量大于背面数量的概率。
很裸的概率DP，我们思考一下状态和转移方程。
我们这样定义状态，定义dp[i][j] 为到第i个硬币时有j个是正面的概率。那么所求答案为sum{ dp[n][i] || (n+1)/2<=i<=n}
题目说了n为odd，
那么状态转移即为： dp[i][j]=dp[i-1][j-1]*p[i]+dp[i-1][j]*(1.0-p[i]);
意思为，到了第i个硬币时，j个正面朝上的状态可以由以下两个状态转移过来：
1、第i-1个硬币的时候，有j-1个正面朝上的，第i个硬币也正面朝上。
2、第i-1个硬币的时候，有j个正面朝上的，第i个硬币反面朝上。
然后初始状态定义

　　

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

注意处理下边界情况就好了。

细节见AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
double p[maxn];
double dp[][];
int n;
int main()
{
    gg(n);
    repd(i,,n)
    {
        scanf("%lf",&p[i]);
    }
    dp[][]=p[];
    dp[][]=1.0000000-p[];

    repd(i,,n)
    {
        for(int j=;j<=i;j++)
        {
            if(j==)
            {
                dp[i][j]=dp[i-][j]*(1.00000-p[i]);
                continue;
            }
            dp[i][j]=dp[i-][j-]*p[i]+dp[i-][j]*(1.0000-p[i]);
//            dp[i][j-1]=dp[i-1][j-1]*(1.000000-p[i]);
        }
    }
    double ans=0.0000000000000000000;
    for(int i=(n+)/;i<=n;i++)
    {
        ans+=dp[n][i];
    }
    printf("%.10lf\n", ans);
    return ;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '');
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  - ch + '';
        }
    }
    else {
        *p = ch - '';
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  + ch - '';
        }
    }
}