hdu4586 Play the Dice 概率 + 等比数列求和 + 极限

Play the Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2914    Accepted Submission(s): 927
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

什么题啊这是，题意感觉不太清晰啊……读着以为每次投的时候，要算有特殊标记的上的数字，然后再投，其实投到有特殊标记的就只是多获得的一次机会而已啊？

并不算这个特殊面上的数字是多少。

这样的话每次投到有特殊标记的概率是m / n

设q = m / n, a = sum / n

那么最后的期望就是

E(Y) = a + a * q + a * q^2 + a * q^3 + ... + a * q^k ,    k->无穷

E(Y) = lim a(1 - q^(k + 1)) / (1 - q),　k ->无穷

这时候就看q = m / n的大小了

如果m == n，那么1 - q == 0，发散

如果m > n，那么q^(k + 1)趋于无穷，发散

如果m < n，那么E(Y) = a(1 - 0) / (1 - q) = (sum / n) / (1 - m / n) = sum / (n - m)

另外就是sum = 0的时候不管怎么样最后肯定是0

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int n, m, sum;
int a[205], b[205];

int main()
{
    while (~scanf("%d", &n)) {
        sum = 0;
        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        scanf("%d", &m);
        for (int i = 0; i < m; i++) {
            scanf("%d", &b[i]);
        }
        if (sum == 0) {
            puts("0.00");
            continue;
        }
        if (m >= n) {
            puts("inf");
        } else {
            printf("%.2f\n", 1.0 * sum / (n - m));
        }
    }
    return 0;
}