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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 ≤ 104
2 ≤ N ≤ 109
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.
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 |
hide comments
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2015-08-29 13:09:40 kartikay singh
easy!!! AC in 1 go |
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2015-08-26 23:35:49 :.Mohib.:
Enjoyed Solving.. nice one... :) |
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2015-08-26 16:29:32 rahul satish
SHAN ROCKS |
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2015-08-26 16:12:48 laksh
got ac:-) Last edit: 2015-08-28 17:29:53 |
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2015-08-25 22:17:42 Advitiya
nice problem :) |
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2015-08-23 19:05:48 anshal dwivedi
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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2015-08-23 18:23:28 priyank
what is output for 1 300 30 is it 299 or -1? |
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2015-08-22 18:09:49 Soumik Chatterjee
Last edit: 2015-08-22 18:14:23 |