P(ξ=K)= C(n,k) * p^k * (1-p)^(n-k), 其中C(n, k) = n!/(k! * (n-k)!)注意！:第二个等号后面的括号里的是上标，表示的是方幂。
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn = 200000 + 5;
long double logF[maxn*2+1];
long double logC(int m,int n){
return logF[m]-logF[n]-logF[m-n];
}
int main()
{
for(int i=1;i<=maxn*2;i++)
logF[i]=logF[i-1]+log(i);
int n,kase=0;
double p;
while(scanf("%d%lf",&n,&p)==2){
double ans=0.0;
for(int i=1;i<=n;i++){
long double c = logC(2*n-i,n);
long double v1=c+(n+1)*log(p)+(n-i)*log(1-p);
long double v2=c+(n+1)*log(1-p)+(n-i)*log(p);
ans +=(double) i*(exp(v1)+exp(v2));
}
printf("Case %d: %.6lf\n",++kase,ans);
}
}
原文：http://blog.csdn.net/a197p/article/details/45574895