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
mkfeuhrer:
2016-06-07 22:45:10
nice problem
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deadbrain:
2016-06-04 14:27:39
what will be the answer for the test case?
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akshayvenkat:
2016-03-15 16:34:37
how to get rid of TLE in this problem ? |
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bubbler9903:
2016-01-29 06:51:50
Cost me a TLE due to the -1 case... |
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rahul2907:
2016-01-19 18:20:03
AC in 1 go :) :D
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wood1234:
2015-10-23 06:43:06
time limit excedded...what to do??answers are correct acording to spoj toolkit... |
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Jaswanth:
2015-09-15 16:44:15
forgot to print -1 costed me 1 wa |
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bka:
2015-09-12 15:32:37
Got It!! Last edit: 2015-09-12 21:12:07 |
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Raghvendra pandey:
2015-09-12 14:59:53
nice problem........ AC :) |
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imeiam:
2015-08-30 20:20:55
Good question. Good logic. |
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 |