hdu 4465 Candy(概率公式+极大乘极小，4级)

最新推荐文章于 2025-08-15 23:10:13 发布

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最新推荐文章于 2025-08-15 23:10:13 发布

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文章标签： java

原文链接：http://www.cnblogs.com/nealgavin/p/3797554.html

这篇文章探讨了一个关于概率和期望值的数学问题，通过一个关于孩子选择糖果盒并吃糖果的情境，来解释如何计算剩余糖果的数量。利用概率论的知识，解决实际问题。

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Candy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1016 Accepted Submission(s): 449
Special Judge

Problem Description

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
  

Source

2012 Asia Chengdu Regional Contest

Recommend

liuyiding

思路：

而前面都可以递推，只有p^(n+1)是常数需要算，但这是一个极小值，而另一坨是个极大值，相乘是个中间值在double内，所以就可以极大极小一点点乘控制在double 范围内就行了，记录乘的个数，不够的补上就好了。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const double nan = 1e-6;
const double inf = 1e6;
const int mm=2e5+9;
double ppp[mm];
int num[mm];
double cal(double p,int n)
{
    double ans = 0, c = 1.0;
    double q = 1.0 - p;
    ppp[0]=1;
    for(int i=1;i<=n+2;++i)
      ppp[i]=ppp[i-1]*p;
    int i;
    int cnt = 0;
    num[0]=0;
    for(i=1; i<=n; i++)
    {
        c *= (double(n+i)/double(i) * q);
       num[i]=num[i-1];
       if(c>inf)
       {
         while(num[i]<=n&&c>nan)
         { num[i]++;
           c*=p;
         }
       }
       ans+=(n-i)*c*ppp[ n+1-num[i] ];
       // ans += (n-i)*c;
    }
    ans += n*ppp[n+1];
//    for(i=0; i<=n; i++)
//        ans *= p;
    return ans;
}
int main()
{
    int n,ca = 0;
    double p;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%lf",&p);
        double ans = 0;
        ans += cal(p,n);
        ans += cal(1-p,n);
        printf("Case %d: %.6lf\n",++ca,ans);
       // cout<<ans<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/nealgavin/p/3797554.html