I ask if i can calculate xor to all numbers in a specific range without using loop
for example if i have
start=3
end =3211233432145321;
the result start ^ start+1 ^ start+2 ^ start+3.........end-1^end
Thanks in advanced.
# | User | Rating |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 160 |
5 | nor | 157 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | pajenegod | 145 |
I ask if i can calculate xor to all numbers in a specific range without using loop
Name |
---|
Finally,
you mention that you need only last 4 numbers so to calculate f(x)
f(x)=(x-3)^(x-2)^(x-1)^(x)
i test it by using this function vs bruteforce solution but the result different.
I wrote C++ code for this: https://github.com/liruqi/topcoder/blob/master/leetcode/exclusive-or-of-numbers-range.cpp