Description
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 numberspi(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 Input
4 0.1 0.2 0.3 0.8
0.800000000000
2 0.1 0.2
0.260000000000
Hint
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.
題意：n個人，每個人提出一個問題的概率爲p，Andrey 可以選擇請求一個人，兩個人，，，，n個人，來得到一個問題，最後要求得到一個問題的最大概率
將n個事件成功的概率從大到小排序，
然後，
只選擇前1件事去做成功的概率就是“從中選擇1件事情去做，最大的實現概率”
只選擇前2件事去做成功的概率就是“從中選擇2件事情去做，最大的實現概率”
以此類推。
證明：選兩種的情況：設a爲其中一個的概率，b爲另一個的概率，則得到一個問題的概率爲a*(1-b)+b*(1-a)=a+b-2*a*b;因爲a,b是小於11的，a*b的增長，遠遠小於a+b的增長，同理，三種，四種情況也是這樣證明的，所以選擇的時候，先從概率從大到小排序，選幾個就選前幾個AC代碼如下：
#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
return *(double*)a>*(double*)b?-1:1;
}
int main()
{
int n,i,j,k;
double f[110],m,m1;
while(scanf("%d",&n)!=EOF)
{
double sum[110]={0.0};
for(i=0;i<n;i++)
{
scanf("%lf",&f[i]);
}
qsort(f,n,sizeof(f[0]),cmp);
sum[0]=f[0];
for(i=1;i<n;i++)//n-1種選法
{
m1=0.0;
for(j=0;j<i+1;j++)//選i+1個
{
m=f[j];
for(k=0;k<i+1;k++)
{
if(k!=j)
m*=(1-f[k]);
}
m1+=m;
}
sum[i]+=m1;
}
qsort(sum,n,sizeof(sum[0]),cmp);
printf("%.12lf\n",sum[0]);
}
return 0;
}