Codeforces442B_Andrey and Problem(贪心)

Andrey and Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.

Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi(0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.

Output

Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.

Sample test(s)input

40.1 0.2 0.3 0.8

output

0.800000000000

input

20.1 0.2

output

0.260000000000

Note

In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.

In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8   0.9·0.2 = 0.26.

解题报告

好久不做cf，大晚上想来水水。。。

看来自己还是太嫩了。

Andrey想让朋友出题，他知道每一个朋友出题的概率，他仅仅想要一道题，而且想要实现的概率最大。

假设Andrey之选一个人问的话一定是概率最大的。选两个人一定是概率前两的人，这就是贪心策略。假设选择概率最大的那个人，实现的概率就是它本身，假设选择前面两大概率P1,P2的，实现的概率就是P1*(1-P2) (1-P1)*P2。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;double num[1000];int cmp(double a,double b){    return a>b;}int main(){    int n;    while(cin>>n)    {        double sum=0,cnt=0,maxx=0;        for(int i=0; i<n; i  )            cin>>num[i];        sort(num,num n,cmp);        for(int k=1; k<=n; k  )        {            sum=0;            for(int i=0; i<k; i  )            {                cnt=1;                for(int j=0; j<k; j  )                {                    if(i!=j)                    {                        cnt*=(double)(1-num[j]);                    }                    else cnt*=num[j];                }                sum =cnt;            }            maxx=max(sum,maxx);        }        printf("%.12lf\n",maxx);    }    return 0;}

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