| # | User | Rating |
|---|---|---|
| 1 | tourist | 3985 |
| 2 | jiangly | 3814 |
| 3 | jqdai0815 | 3682 |
| 4 | Benq | 3529 |
| 5 | orzdevinwang | 3526 |
| 6 | ksun48 | 3517 |
| 7 | Radewoosh | 3410 |
| 8 | hos.lyric | 3399 |
| 9 | ecnerwala | 3392 |
| 9 | Um_nik | 3392 |
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 169 |
| 2 | maomao90 | 162 |
| 2 | Um_nik | 162 |
| 4 | atcoder_official | 161 |
| 5 | djm03178 | 158 |
| 6 | -is-this-fft- | 157 |
| 7 | adamant | 155 |
| 8 | awoo | 154 |
| 8 | Dominater069 | 154 |
| 10 | luogu_official | 150 |
You are given two arrays a and b each of size n. For each i (0<=i<n), you have to find smallest 'x' such that
-> x+y = a[i]
-> x^y = b[i] (i.e. x xor y = b[i])
0<=a[i],b[i]<=10^(15)
1<=n<=50
It is possible that there might be no 'x' possible. Would be glad if anyone could help...
It is possible
By x^y do you mean power or the bitwise operator?
Use this property :- (x + y) = (x ^ y) + 2*(x & y)