uva1639 Candy

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LazyChild is a lazy child who likes candy very much. Despite being
very young, he has two large candy boxes, each contains n candies
initially. Everyday he chooses one box and open it. He chooses the
rst box with probability p and the second box with probability (1

不妨设最后一个打开的是第一个盒子【最后打开的是第二个同理】，第二个盒子还剩i个，那么对应的数学期望为C(2 * n-i,n) * p^(n+1)【最后又打开一次】 * (1-p)^(n-i)*i。
这个式子是对的，但是有两个问题，一个是C太大，存不下，二是p^n太小，有精度误差。
所以可以利用对数，计算i*e^((n+1) * ln(p)+(n-i) * ln(1-p)+lnfac[2 * n-i]-lnfac[n]-lnfac[n-i])，其中lnfac[x]表示ln(x!)，可以预处理出来。这样计算过程中就不会出现太大或者太小的数了。
还要注意，这样对精度要求还是不够的，需要用到long double。

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<iomanip>using namespace std;#define LD long doubleconst LD eps=1e-14;LD lnfac[400010];int main(){    int i,j,k,m,n,K=0;    LD x,y,z,p,q,ans;    for (i=2;i<=400000;i++)      lnfac[i]=lnfac[i-1]+log(i);    while (cin>>n>>p)    {        if (p<=eps||fabs(1-p)<=eps)        {            printf("Case %d: %d.000000\n",++K,n);            continue;        }        ans=0;        for (i=1;i<=n;i++)          ans+=i*(exp((n+1)*log(p)+(n-i)*log(1.0-p)+lnfac[2*n-i]-lnfac[n]-lnfac[n-i])              +exp((n+1)*log(1.0-p)+(n-i)*log(p)+lnfac[2*n-i]-lnfac[n]-lnfac[n-i]));        cout.setf(ios::fixed);        cout<<"Case "<<++K<<": "<<fixed<<setprecision(6)<<ans<<endl;    }}

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