Play the Dice（HDU OJ 4586）

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#HDUOJ 4586 #c++ #等比数列求和公式 #模拟与化简

解析PlaytheDice游戏规则，玩家投掷一个多面骰子并根据出现的面获得相应金额，某些特殊面允许额外投掷一次。文章提供了一种算法来计算玩家参与该游戏一次的预期收益。

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Play the Dice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1297 Accepted Submission(s): 415
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

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代码：

/******************************
*   @name:hdu oj 4586         *
*   @author:npufz             *
*******************************/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
using namespace std;
int main()
{
    int  a,n,m,sum;
  while(cin>>n)
  {
      sum=0;
      for(int i=0;i<n;i++)
      {
          cin>>a;
          sum+=a;
      }
      cin>>m;
      for(int i=0;i<m;i++)
      {
          cin>>a;
      }
      if(sum==0) printf("0.00\n");
      else
      {
          if(n==m) printf("inf\n");
          else
          {
              printf("%.2lf\n",1.0*sum/(n-m));
            }
      }
  }
return 0;
}

反思：一开始没有对等比数列进行运用求和公式，先WA后TLE,后来用了公式还是WA了，因为没有判断N==M时，如果每个面的数字为零的话和为零而不是INF