uva - 10833 Supermean(二项式系数，对指数)

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