Candy HDU - 4465 （数学）（概率期望）

最新推荐文章于 2019-01-11 01:00:00 发布

原创最新推荐文章于 2019-01-11 01:00:00 发布 · 447 阅读

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本文介绍了一道关于懒小孩吃糖的概率题目，详细解释了如何通过概率论和组合数学的方法来计算当一个喜欢糖果的小孩吃完一盒糖后，另一盒中剩余糖果的期望数量。文章给出了具体的算法实现，并提供了示例输入输出。

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 first box with probability p and the second box with probability (1 - p). For the chosen box, if there are still candies in it, he eats one of them; otherwise, he will be sad and then open the other box.
He has been eating one candy a day for several days. But one day, when opening a box, he finds no candy left. Before opening the other box, he wants to know the expected number of candies left in the other box. Can you help him?
Input
There are several test cases.
For each test case, there is a single line containing an integer n (1 ≤ n ≤ 2 × 10 5) and a real number p (0 ≤ p ≤ 1, with 6 digits after the decimal).
Input is terminated by EOF.
Output
For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is a real number indicating the desired answer.
Any answer with an absolute error less than or equal to 10 -4 would be accepted.
Sample Input
10 0.400000
100 0.500000
124 0.432650
325 0.325100
532 0.487520
2276 0.720000
Sample Output
Case 1: 3.528175
Case 2: 10.326044
Case 3: 28.861945
Case 4: 167.965476
Case 5: 32.601816
Case 6: 1390.500000

这题在黑书上是有的，但是我给忘了，这个专题开辟的太晚了，只能希望在比赛的时候不要遇到，但是今天就认真的记录一下吧
那么我们看有多少种情况，假设当停止摸糖时，剩下 $i$ 颗糖
那么，一共操作了 $n+n-i$
假设这i颗糖在第二个盒子里
则剩下i颗的概率是：

C n 2 n - i p n + 1 (1 - p) n - i

$C_{2n-i}^np^{n+1}(1-p)^{n-i}$
同理在第一个盒子里剩下：

C n 2 n - i (1 - p) n + 1 p n - i

$C_{2n-i}^n(1-p)^{n+1}p^{n-i}$
因为精度问题，得取个对数（学到了）:

p (i) = e l n C n 2 n - i + (n + 1) l n p + (n - i) l n (1 - p) + e l n C n 2 n - i + (n + 1) l n (1 - p) + (n - i) l n p

$p(i)=e^{lnC_{2n-i}^n+(n+1)lnp+(n-i)ln(1-p)}+e^{lnC_{2n-i}^n+(n+1)ln(1-p)+(n-i)lnp}$
然后

a n s = \sum i = 1 n p (i) * i

$ans=\sum_{i=1}^{n}p(i)*i$

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 using namespace std;
 const int maxn=400040;
 double fac[maxn],p,q; 
 int n;
 double cal(int n,int m)
 {
     return fac[n]-fac[m]-fac[n-m];
 }
 int main()
 {
     int cas=0;
     memset(nlog,0,sizeof(nlog));
     for(int i=1;i<maxn;i++)
        fac[i]=fac[i-1] + log(i);
     while(scanf("%d%lf",&n,&p)!=EOF)
     {
         q=log(1-p); 
         p=log(p);
         double ans=0;
         for(int i=1;i<=n;i++)
         {
             double v1=cal(2*n-i,n)+(n+1)*p+(n-i)*q;
             double v2=cal(2*n-i,n)+(n+1)*q+(n-i)*p;
             ans+=i*(exp(v1)+exp(v2));
         }
         printf("Case %d: %.6lf\n",++cas,ans);
     }
     return 0;
 }