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:

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <deque>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <iomanip>

using namespace std;

#define INF 0xffffff7
#define maxn 50010
const double tmp = log(2.0);
double data[maxn];
int main()
{
    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        int n;
        scanf("%d", &n);
        double ans = 0.0, c = 0.0;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf", &data[i]);
            if(data[i] > 0) ans += exp(log(data[i]) - (n-1)*log(2.0) + c);
            else if(data[i] < 0) ans -= exp(log(-data[i]) - (n-1)*log(2.0) + c);
            //cout << ans << endl;
            c += log((double)n-i-1)-log((double)i+1);
        }
        printf("Case #%d: %.3lf
", kase, ans);
    }
    return 0;
}