| # | User | Rating |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3773 |
| 3 | Radewoosh | 3646 |
| 4 | ecnerwala | 3624 |
| 5 | jqdai0815 | 3620 |
| 5 | Benq | 3620 |
| 7 | orzdevinwang | 3612 |
| 8 | Geothermal | 3569 |
| 8 | cnnfls_csy | 3569 |
| 10 | gyh20 | 3447 |
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 161 |
| 1 | Um_nik | 161 |
| 3 | maomao90 | 160 |
| 4 | awoo | 159 |
| 5 | atcoder_official | 157 |
| 6 | -is-this-fft- | 156 |
| 7 | nor | 155 |
| 7 | adamant | 155 |
| 9 | maroonrk | 152 |
| 10 | Dominater069 | 148 |
Problem: Given an array a1, a2, ..., an (1 <= n <= 1e6) and q queries, for each query has two types.
1 l r x: a[i] = min(a[i], x) (l <= i <= r).
2 i : print value in position i.
I dont know use lazy update to solve operator 1, help me pls (sorry i'm poor E).
hi everyone!
I read a blog write about formula of a^b = a'b + ab' and i dont understand. Someone can explain it for me pls :((
(sorry i'm poor E).
link: https://www.homeworklib.com/qaa/1383353/2prove-that-a-xor-b-a-xor-c-a-xor-b-b-xor
