标签：style blog color os io for div sp 代码
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\n", kase, ans); 43 } 44 return 0; 45 }
uva - 10833 Supermean(二项式系数，对指数)
标签：style blog color os io for div sp 代码
原文地址：http://www.cnblogs.com/LLGemini/p/3958634.html
