BITPLAY - PLAYING WITH BITS
The problem is very simple.
You are given a even number N and an integer K and you have to find the greatest odd number M less than N such that the sum of digits in binary representation of M is at most K.
Input
For each test case, you are given an even number N and an integer K.
Output
For each test case, output the integer M if it exists, else print -1.
Constraints
1 <= T <= 10^4
2 <= N <= 10^9
0 <= K <= 30
Example
Input: 2 10 2 6 1 Output: 9 1
Explanation
First case when N = 10, K = 2
Binary representation of 10 is 1010 and binary representation of 9 is 1001, hence greatest odd number less than 10 whose sum of digits in its binary representation is at most 2 is 9. Hence output is 9.
hide comments
kartikay singh:
2015-08-29 13:09:40
easy!!! AC in 1 go |
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:.Mohib.::
2015-08-26 23:35:49
Enjoyed Solving.. nice one... :)
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rahul satish:
2015-08-26 16:29:32
SHAN ROCKS |
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laksh:
2015-08-26 16:12:48
got ac:-) Last edit: 2015-08-28 17:29:53 |
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Advitiya:
2015-08-25 22:17:42
nice problem :) |
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anshal dwivedi:
2015-08-23 19:05:48
it's 299@priyank ..!by the way very easy problem but don't forget to print -1, got 1 WA just because of that..!:) |
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priyank:
2015-08-23 18:23:28
what is output for
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Soumik Chatterjee:
2015-08-22 18:09:49
Last edit: 2015-08-22 18:14:23 |
Added by: | tapariaankit |
Date: | 2015-08-22 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 GOSU JS-MONKEY |