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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 <= 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.
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 |
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2020-05-20 04:03:35
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 " exactly " K ?? if you change "atmost" become "exactly"! it will be interesting! Last edit: 2020-05-20 04:03:59 |
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2019-04-02 14:44:09
AC in one go:) nice problem. |
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2019-01-20 15:37:59
is there necessary to check k<=30 or not . im getting answer according to the given testcases and also checking -1 condition still giving me wrong answers any one has idea Last edit: 2019-01-20 15:40:29 |
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2018-08-11 21:55:08
I feel so confident solving it without the bitset library lol |
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2017-07-04 11:44:52
bitset rocks!!...dont forget that -1 |
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2017-06-03 23:30:07
use bitset libraray functions |
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2017-05-09 18:04:19
very easy logic ;P |
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2017-02-01 20:40:10
nice problem learnt bitset library |
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2016-12-23 07:13:33
poor -1 costed me 2 wa... anyways nice question... |
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2016-08-11 12:29:09
"Atmost k " COSTED ME 1 WA |