Microsoft 2017 Foundation program online programming pen test legendary Items

Title 1:legendary Items time limit: 10000ms single point limit: 1000ms memory limit: 256MB description

Little Hi is playing a video game. Accomplishes a quest in the game and 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'll go up q%. Since the probability is getting higher he'll get a legendary item eventually.

After getting a legendary item the probability is 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 would also go up q% each time Little Hi accomplishes a quest until he gets another item.

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

Assume P = 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

Input

The contains three integers P, Q and N.

1≤n≤106, 0≤p≤100, 1≤q≤100 output

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

50 75 2

Sample output

3.25

Read for a long time to understand.

The probability of completing a quest with no legendary item is increased q% (until legendary item is obtained), legendary item quest, and the probability is reset to p/(2^i)%

Reflected in the diagram is: the left node of each node is a solid node (completed legendary item), the right node is an empty node (legendary item is not completed). The left node of each empty node (assuming probability x) is X+q, and the left node probability of the solid node is p/(2^i)%, (the sum of the right and left nodes of each node in the graph is 100%). How much quest is needed to get n a Legendaryitem

The above figure for example, from the top down, p is 50%,n for 2 o'clock, left to go, the probability is: 50%,25% (2 legendary item), then stop

View the entire tree as an n-tier, starting at the bottom.

Here's the calculation: from the bottom up, starting with the leftmost 25% node, this is 0.25* (0+1), and you don't have to go down (because you've got 2 legendary item)

The right node is 1-0.25=0.75 (former), downward continuation probability is 0.25+q=1 (if less than 1, can go down), the probability is 0.75* (0+2). The sum of the two points and 0.25+0.75*2=1.75

The parent node of 25% is 50%, then 50% is 0.5* (1.75+1), meaning that the child node of this point plus 1 level multiplied by 0.5 is this point value

Look at the other 50% (right node), former 0.5, continue down to 0.5* (1.75+2), two points and 3.25

AC Code:

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 >) {
    				j = n;
    			}
    			One + + former * (double) j/100 * (r + level);
    			Former *= (1.0-(double) j/100.0);
  
    			level++;
    			if (j = =) {break
    			    ;
    			}
    		}
    		R = one;
    	}
    	System.out.print (R);
	}

And then recommend a link to explain the written test, the analysis is very good:

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