Given the xor of n numbers,I need to find those n numbers (all positive) such that their sum is minimum. I thought on the lines of minimising the bitwise and of the n numbers, but couldn't come up with a solution. Thanks in advance
# | User | Rating |
---|---|---|
1 | jiangly | 3898 |
2 | tourist | 3840 |
3 | orzdevinwang | 3706 |
4 | ksun48 | 3691 |
5 | jqdai0815 | 3682 |
6 | ecnerwala | 3525 |
7 | gamegame | 3477 |
8 | Benq | 3468 |
9 | Ormlis | 3381 |
10 | maroonrk | 3379 |
# | User | Contrib. |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | Dominater069 | 161 |
4 | Um_nik | 160 |
5 | atcoder_official | 159 |
6 | djm03178 | 157 |
7 | adamant | 153 |
8 | luogu_official | 150 |
9 | awoo | 149 |
10 | TheScrasse | 146 |
Given the xor of n numbers,I need to find those n numbers (all positive) such that their sum is minimum. I thought on the lines of minimising the bitwise and of the n numbers, but couldn't come up with a solution. Thanks in advance
Name |
---|
First break the given xor number like if the number is 11 then break it into 1 + 2 + 8 let's say we store these numbers into a vector arr.
Now if n is less than no. of set bits in the given xor number . Then we need to merge some numbers and it can be done easily.
In other case let's define variable rem = n — no. of set bits in given xor number .
Now if rem is even then we can make all remaining number 1 .
If rem is odd then we find a even number from the arr and add 1 to it . Now fill all the rem no. with 1 this is gonna work (think why ?).
There may be some corner cases when given xor no. is 1 .(Think how can you handle this ) .
Sorry I am very bad at using CF features .