| # | 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 |
we are given 4 integers, where a <= b , c <= d.
We have to find Sum of Xor of all the pairs (i,j) such that ( a <= i <= b , && c <= j <= d )
int sum = 0;
for(int i=a ; i <= b; i++) {
for(int j = c; j <= d ; j++) {
sum += (i ^ j)
}
}
return sum;
How to find this sum optimally ?
For each x calculate the effect of 2^x on the answer.
The problem then becomes 32 "at position i, calculate how many zeros are in a certain interval". After getting the number of 1's and 0's just multiply them together. Then add up all the answers and you're done.