hdu4586 Play the Dice 概率 + 等比數列求和 + 極限

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

什麼題啊這是，題意感覺不太清晰啊……讀着以爲每次投的時候，要算有特殊標記的上的數字，然後再投，其實投到有特殊標記的就只是多獲得的一次機會而已啊？

並不算這個特殊面上的數字是多少。

這樣的話每次投到有特殊標記的概率是m / n

設q = m / n, a = sum / n

那麼最後的期望就是

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

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

這時候就看q = m / n的大小了

如果m == n，那麼1 - q == 0，發散

如果m > n，那麼q^(k + 1)趨於無窮，發散

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

另外就是sum = 0的時候不管怎麼樣最後肯定是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;
}