「CF442B」Andrey and Problem
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.
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.
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.
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1 2 |
4 0.1 0.2 0.3 0.8 |
|
1 |
0.800000000000 |
|
1 2 |
2 0.1 0.2 |
|
1 |
0.260000000000 |
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.
题解
证明我不懂啊。。。
直觉告诉我们排序后从大到小如果取完更优就取，不然输出答案
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#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #define inf 0x7fffffff #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m; double a[105],ans,s; int main() { n=read(); for(int i=1;i<=n;i++) scanf("%lf",&a[i]); sort(a+1,a+n+1); ans=a[n];s=(1-a[n]); for(int i=n-1;i>0;i--) { double t=ans*(1-a[i])+a[i]*s; if(t>ans)ans=t; else break; s*=(1-a[i]); } printf("%.12lf",ans); return 0; } |
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