http://acm.hdu.edu.cn/showproblem.php?pid=4465

第一直觉概率DP但很快被否定，发现只有一个简单的二项分布，但感情的表达，没有对生命和死亡的例子。然后找到准确的问题，将不被处理，

事实上与思考C递归成为O(1)。每次乘以p

代码看这里http://fszxwfy.blog.163.com/blog/static/44019308201338114456115/

贴一个我没看懂的代码

//#pragma comment(linker, "/STACK:102400000,102400000")

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <iostream>

#include <iomanip>

#include <cmath>

#include <map>

#include <set>

#include <queue>

using namespace std;

#define ls(rt) rt*2

#define rs(rt) rt*2+1

#define ll long long

#define ull unsigned long long

#define rep(i,s,e) for(int i=s;i<e;i++)

#define repe(i,s,e) for(int i=s;i<=e;i++)

#define CL(a,b) memset(a,b,sizeof(a))

#define IN(s) freopen(s,"r",stdin)

#define OUT(s) freopen(s,"w",stdout)

const ll ll_INF = ((ull)(-1))>>1;

const double EPS = 1e-8;

const double pi = acos(-1.0);

const int INF = 100000000;

double p1,p0;

int n;

double solve()

{

    p1=1.0-p0;

    double tmpa,tmpb;

    tmpa=tmpb=1.0;

    double ans=0.0;

    int last=n+1;

    for(int i=0;i<n;i++)

    {

        if(i)

        {

            tmpa=tmpa*(n+i)*p0/i;

            tmpb=tmpb*(n+i)*p1/i;

            while(tmpa>n || tmpb>n)

            {

                tmpa*=p1;

                tmpb*=p0;

                last--;

            }

        }

        ans+=(n-i)*tmpa*pow(p1,last)+tmpb*(n-i)*pow(p0,last);

    }

    return ans;

}

int main()

{

    //IN("hdu4465.txt");

    int ic=0;

    while(~scanf("%d%lf",&n,&p0))

    {

        printf("Case %d: %.6lf\n",++ic,solve());

    }

    return 0;

}