| № | Пользователь | Рейтинг |
|---|---|---|
| 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 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 165 |
| 3 | adamant | 161 |
| 4 | TheScrasse | 160 |
| 5 | nor | 158 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | orz | 146 |
| 9 | SecondThread | 145 |
| 9 | pajenegod | 145 |
Given an array of integers $$$(A[i] <= 10^5)$$$. We have to find number of unordered pairs (i, j) such that their GCD is greater than B.
All value <= $$$10^5$$$
Can someone give me a general idea how to approach such kinds of problems? Because this type of question appears frequently in programming contests and every time I devote much time solving this but failed each time :(
You can work backward, let dp[i] be the ammount of pairs that have gcd equals to i in vector, for a value A[j] we know that the gcd(A[j], y) = d such that d is a divisor of A[j], so lets make freq[i] = ammount of values that have i as divisor. dp[i] = (freq[i] choose 2) — dp[2*i] — dp[3*i] — ... — dp[k*i] We can fill dp array in nlog(n), the answer is the suffix-sum of dp array
I am thinking in this way only. But failed to land on this solution.
Do you know any blog or some questions similar to this?
D. The Millennium Prize Problems