SCPC11G - Indomie

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During recession, Amjad needs to queue for SembakoPlus. Sembako, as we all know, stands for “Sembilan Bahan Pokok” which consists of 9 kinds of item: Rice, Sugar, Cooking-oil, Meat, Egg, Milk, Corn, Kerosene and Iodized Salt. SembakoPlus consists of Sembako and one more item: Indomie! Amjad’s favorite of all time!! (therefore, no wonder why he could stand for this long queue).

Each person in the queue is allowed to pick only one item. No need to ask, Amjad wants only Indomie. Unfortunately, they are running out of SembakoPlus stock and currently there are three kinds of item left: Rice, Sugar and Indomie. As he could see from afar, he is quite sure that Rice and Sugar will be enough for everybody.

Given the number of remaining Indomie and the number of people queuing in front of Amjad, your task is to count the probability that he will get his Indomie. Amjad can’t do programming right now as he is very nervous so he can’t think logically. He needs your help!

 

 

Input

 

There will be multiple test cases for this problem. Each test case contains two integers N (1 ≤ N ≤ 50) and S (0 ≤ S ≤ 50), where N is the number of people queuing in front of Amjad and S is the remaining number of Indomie.

Output

 

 

For each case, print in a single line the probability in percentage that he will get his Indomie with 5 digits precision (he’s being paranoid)

 

Example

Input:
2 1
3 2
4 0
4 1
10 10
14 9
30 14


Output:
50.00000
76.92308
0.00000
33.33333
99.99831
98.65515
95.16071
Explanation for 1st sample test case:
There are two peoples queuing in front of Amjad, so those two peoples could pick of the
following combination {1st people, 2nd people}:
1. Rice, Rice
2. Rice, Sugar
3. Rice, Indomie
4. Sugar, Rice
5. Sugar, Sugar
6. Sugar, Indomie
7. Indomie, Rice
8. Indomie, Sugar
Since there is only one Indomie left, there are only 4 out of 8 combinations that ensure Amjad to get his Indomie (1, 2, 4 and 5), hence the probability is 4/8 = 50%.

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hodobox: 2017-05-07 13:42:05

Yeah, the probabilities of each sequence don't follow a probability tree. You are just supposed to count (number of possible sequences where at least 1 Indomie is not eaten)/(number of all possible sequences)

Last edit: 2017-05-07 13:42:48
zicowa: 2014-07-14 17:17:34

A simple optimisation will definately work :P :D O((n-k)*k) passed AC

Hussain Kara Fallah: 2012-08-20 01:31:05

You should use BIG INT .... Be careful !

Bharath Reddy: 2012-07-11 09:44:20

@fitcat and Tywan: You can't have the combination Indomie, Indomie for the first test case. So the probability is only 0.5

berney: 2012-03-26 08:50:33

my code runs fine in ideone(in java6)...
but here it is giving compilation error...
can anyone please tell what is wrong?
solution id 6726890...

Devil D: 2011-12-28 08:45:03

what is the answer for 20 19 ?
100.00000?

I am getting correct result for the given test cases. but WA here. my submission is '6255319'

Last edit: 2011-12-28 08:48:36
fitcat: 2011-11-26 09:27:47

The example output is incorrect. For the 1st sample test case, the correct answer should be 2/3*2/3 = 4/9 = 44.44444 as Tywan pointed it out.

Tywan: 2011-10-27 08:13:58

I've solved the problem but still believe the right answer should be
44.44444
74.07407
and so on...

Lukmanul Hakim: 2011-10-24 01:46:27

What? Indomie from Syria? :O

Ayan Ghatak: 2011-10-21 13:59:07

how long do i scan??till eof??


Added by:Ali Arous
Date:2011-10-19
Time limit:0.903s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:Syrian Collegiate Programming Contest 2011