HDU 4586（数学期望）

最新推荐文章于 2020-08-04 23:38:19 发布

原创最新推荐文章于 2020-08-04 23:38:19 发布 · 656 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#math #期望

Math 专栏收录该内容

9 篇文章

订阅专栏

探讨了一款特殊投骰子游戏的数学期望收益计算方法，通过分析不同颜色的骰子面及其对应的再次投掷机会，给出了计算公式及代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Play the Dice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1830 Accepted Submission(s): 591
Special Judge

Problem Description

There is a dice with n sides, which are numbered from 1,2,...,n and have the equal possibility to show up when one rolls a dice. Each side has an integer ai on it. Now here is a game that you can roll this dice once, if the i-th side is up, you will get ai yuan. What's more, some sids of this dice are colored with a special different color. If you turn this side up, you will get once more chance to roll the dice. When you roll the dice for the second time, you still have the opportunity to win money and rolling chance. Now you need to calculate the expectations of money that we get after playing the game once.

Input

Input consists of multiple cases. Each case includes two lines.
The first line is an integer n (2<=n<=200), following with n integers a _i(0<=a _i<200)
The second line is an integer m (0<=m<=n), following with m integers b _i(1<=b _i<=n), which are the numbers of the special sides to get another more chance.

Output

Just a real number which is the expectations of the money one can get, rounded to exact two digits. If you can get unlimited money, print inf.

Sample Input

  
   6 1 2 3 4 5 6
0
4 0 0 0 0
1 3

Sample Output

  
   3.50
0.00

Source

2013 ACM-ICPC南京赛区全国邀请赛——题目重现

题意：

有一个骰子有n个面，每个面都有一个数字，分别是1~n。现让你投一次骰子，你可以得到骰子朝上的那个面的数字的金额。但有m个面中的数字有些特别，因为假如你投的骰子朝上的那个面是那些数字的话，你不仅可以得到那个面的数字的金额，而且还可以再投一次。问你玩一次这样的游戏能得到的金额的期望是多少？

分析：

第一次投到每个数字的几率都是一样的：1/n，所以第一次期望是sum/n。然后有可能投到那些特别的数字，这样就有第二次机会，第二次的期望是sum*m/n（投到特别数字的几率是m/n，sum为数字总和）。类似的第三次的期望：sum*(m/n)*(m/n)。然后假设总共投了k次，根据期望的可加性：总的期望 E = sum/n*(1+m+m*m/n+m*(m/n)^2...) 里面共k项，它是等比数列，根据等比数列求和公式可以求得(中间还利用了极限理论)：E = sum/(n-m)。一开始我没推出公式，于是直接写了个循环判断精度，然后各种WA==

本题还需注意的一个地方就是即使m == n也就是骰子所有面的数都是特殊的数时answer也不一定是inf，因为所有面的值都为0的情况answer就是0，而不是inf。。

code:

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
    int n, m, a, b;

    while(~scanf("%d", &n))
    {
        double sum = 0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a);
            sum += a;
        }
        scanf("%d", &m);
        for(int i=1; i<=m; i++)
            scanf("%d", &b);
        if(sum == 0) {printf("0.00\n"); continue;} //没写WA
        if(m == n) {printf("inf\n"); continue;}
        printf("%.2f\n", sum/(n-m));
    }
    return 0;
}

由于马上要去上海邀请赛了，所以最近经常模拟比赛。上一场的比赛题目还没搞懂，下一场比赛就来了==。现在很多题也都不愿写题解了，感觉挺麻烦，，还是看心情吧==