微软2017年预科生计划在线编程笔试题Legendary Items

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题目1 : Legendary Items

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Little Hi is playing a video game. Each time he accomplishes a quest in the game, Little Hi has a chance to get a legendary item.

At the beginning the probability is P%. Each time Little Hi accomplishes a quest without getting a legendary item, the probability will go up Q%. Since the probability is getting higher he will get a legendary item eventually.

After getting a legendary item the probability will be reset to ⌊P/(2I)⌋% (⌊x⌋ represents the largest integer no more than x) where I is the number of legendary items he already has. The probability will also go up Q% each time Little Hi accomplishes a quest until he gets another legendary item.

Now Little Hi wants to know the expected number of quests he has to accomplish to get N legendary items.

Assume P = 50, Q = 75 and N = 2, as the below figure shows the expected number of quests is

2*50%*25% + 3*50%*75%*100% + 3*50%*100%*25% + 4*50%*100%*75%*100% = 3.25

输入

The first line contains three integers P, Q and N.

1 ≤ N ≤ 106, 0 ≤ P ≤ 100, 1 ≤ Q ≤ 100

输出

Output the expected number of quests rounded to 2 decimal places.

样例输入

50 75 2

样例输出

3.25

看了半天才看懂。

题意：每次完成一个没有legendary item的quest，概率会上升Q%(直到获得legendary item)；完成legendary item的quest，概率重设为P/(2^I)%

反映到图中就是：每个节点的左节点为实心节点（完成legendary item的），右节点为空节点（未完成legendary item）。每个空节点（假设概率x）的左节点为x+q，实心节点的左节点概率为P/(2^I)% ，（图中每个节点的左右节点之和为100%）。求为了得到n个legendaryitem，需要多少quest

以上图为例，从上往下看，p为50%，n为2时，最左边一直走，概率分别为：50%，25%（2个legendary item），则停止

将整个树看作n层，从底层开始计算。

下面开始计算：自底向上，从最左的25%节点开始，这一点为0.25*（0+1），此时不需往下走(因为已经拿到2个legendary item)

右节点为1-0.25=0.75（former），向下继续走概率为0.25+q=1（如果小于1，可以往下继续走），则概率为0.75*（0+2）。至此两点之和0.25+0.75*2=1.75

25%的父节点为50%，则50%这一点为0.5*（1.75+1），意思是这一点的子节点加上1个level再乘以0.5就是这一点值

再看另外50%（右节点），former为0.5，继续向下走为0.5*（1.75+2），至此两点之和3.25

AC代码：

public class Main {public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int p = in.nextInt();    int q = in.nextInt();    int n = in.nextInt();    helper(p, q, n);}public static void helper(int p, int q, int n) {List<Integer> list = new ArrayList<Integer>();    list.add(p);    double r = 0;    for (int i = 1; i < n; i++) {    list.add(list.get(i - 1) / 2);    }    for (int m = n - 1; m >= 0; m--) {    int tmp = list.get(m);    double one = 0, former = 1;    int level = 1;    for (int j = tmp; ; j+= q) {    if (j > 100) {    j = 100;    }    one += former * ((double)j / 100)* (r + level) ;    former *= (1.0 - (double)j / 100.0);      level++;    if(j == 100) {        break;    }    }    r = one;    }    System.out.print(r);}}

再推荐一个这场笔试的讲解链接，分析的很好：

https://github.com/hiho-coder/msft-2017-online-test-solution/blob/master/README.md

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