Play the Dice - HDU 4586 期望dp

最新推荐文章于 2017-10-31 11:09:43 发布

提比-我有特殊的AC技巧

最新推荐文章于 2017-10-31 11:09:43 发布

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分类专栏： dp动态规划和递推 HDU

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探讨一个特殊掷骰子游戏的数学模型，通过计算不同情况下的收益期望，解决何时可以获得额外掷骰机会的问题。

Play the Dice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1689 Accepted Submission(s): 548
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

题意：一个骰子有n个面，每个面有一个权值，掷到m个面时可以继续掷，问得到的权值的期望是多少。

思路：ans=不能继续掷：ai/n + 继续掷：(ai+ans)/n ，最后结果是ans=(a1+a2+...+an)/(n/m)。再特判inf的情况。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int T,t,n,m,sum;
int main()
{
    int i,j,k;
    double ans;
    while(~scanf("%d",&n))
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&k);
            sum+=k;
        }
        scanf("%d",&m);
        for(j=1;j<=m;j++)
           scanf("%d",&k);
        if(sum==0)
           printf("0\n");
        else if(n==m)
          printf("inf\n");
        else
        {
            ans=round(100.0*sum/(n-m));
            printf("%.2f\n",ans/100);
        }
    }
}