B. Andrey and Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAndrey 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.
InputThe 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.
OutputPrint 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)Input4 0.1 0.2 0.3 0.8Output0.800000000000Input2 0.1 0.2Output0.260000000000NoteIn 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.
Accepted Code:
1 /************************************************************************* 2 > File Name: 442B.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年06月23日 星期一 14时19分19秒 6 > Propose: 7 ************************************************************************/ 8 9 #include10 #include 11 #include 12 #include 13 #include 14 #include 15 #include 16 using namespace std;17 18 #define eps 1e-919 int n;20 double p[105];21 22 double cal(int x) {23 double result = 0.0;24 for (int i = x; i < n; i++) {25 double tmp = p[i];26 for (int j = x; j < n; j++) {27 if (i == j) continue;28 tmp *= (1.0 - p[j]);29 }30 result += tmp;31 }32 return result;33 }34 35 int36 main(void) {37 while (~scanf("%d", &n) && n) {38 for (int i = 0; i < n; i++) scanf("%lf", p+i);39 sort(p, p + n);40 double ans = p[n-1];41 for (int i = n-2; i >= 0; i--) {42 double tmp = cal(i);43 if (tmp - ans < eps) break;44 ans = tmp;45 }46 printf("%.17f\n", ans);47 }48 49 return 0;50 }