uva

模拟发现，每个元素求和时，元素的系数是二项式系数，于是ans=sum(C(n-1,i)*a[i]/2^(n-1)),但是n太大，直接求会溢出，其实double的范围还是挺大的，所以可以将组合数转化成对数:

e^(lnC(n-1, k)*A[k]/(2^n-1) ) ==> e^( ln C(n-1,k) + ln A[k] - (n-1)*ln2 );

又直接利用公式求二项式系数：C(n, k+1)/C(n, k) = (n-k)/(k+1);

而且对数还有递推求法:

logC(n,k+1)=logC(n,k)+log(n-k)-log(k+1)

代码:

 1 #include <iostream>
 2 #include <sstream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <string>
 7 #include <vector>
 8 #include <set>
 9 #include <cctype>
10 #include <algorithm>
11 #include <cmath>
12 #include <deque>
13 #include <queue>
14 #include <map>
15 #include <stack>
16 #include <list>
17 #include <iomanip>
18 
19 using namespace std;
20 
21 #define INF 0xffffff7
22 #define maxn 50010
23 const double tmp = log(2.0);
24 double data[maxn];
25 int main()
26 {
27     int T;
28     scanf("%d", &T);
29     for(int kase = 1; kase <= T; kase++)
30     {
31         int n;
32         scanf("%d", &n);
33         double ans = 0.0, c = 0.0;
34         for(int i = 0; i < n; i++)
35         {
36             scanf("%lf", &data[i]);
37             if(data[i] > 0) ans += exp(log(data[i]) - (n-1)*log(2.0) + c);
38             else if(data[i] < 0) ans -= exp(log(-data[i]) - (n-1)*log(2.0) + c);
39             //cout << ans << endl;
40             c += log((double)n-i-1)-log((double)i+1);
41         }
42         printf("Case #%d: %.3lf
", kase, ans);
43     }
44     return 0;
45 }