hdu 4586 Play the Dice （概率+等比数列）

Play the Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 1328    Accepted Submission(s): 429

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

 

思路：刚開始理解错题意了，老感觉那m个面的数字的大小和结果有关。

事实上，掷到一个特殊面仅仅是得到了一个再次投掷的机会，和第一次投掷的效果全然一样。

sum=a1+a2+...+an; a=sum/n;  q=m/n;

ans=a+a*q+a*q^2+...=sum/(n-m);

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 205
const int inf=0x3fffffff;
int a[N];
int main()
{
    int i,n,m,t;
    double sum;
    while(scanf("%d",&n)!=-1)
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%d",&t);
        }
        if(sum==0)
        {
            printf("0.00\n");
            continue;
        }
        if(m==n)
        {
            printf("inf\n");
            continue;
        }
        printf("%.2f\n",sum/(n-m));
    }
    return 0;
}