| # | User | Rating |
|---|---|---|
| 1 | tourist | 3892 |
| 2 | jiangly | 3797 |
| 3 | orzdevinwang | 3706 |
| 4 | jqdai0815 | 3682 |
| 5 | ksun48 | 3588 |
| 6 | ecnerwala | 3557 |
| 7 | Ormlis | 3532 |
| 8 | Benq | 3468 |
| 9 | Radewoosh | 3463 |
| 10 | Um_nik | 3450 |
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 165 |
| 2 | Qingyu | 159 |
| 3 | -is-this-fft- | 158 |
| 3 | atcoder_official | 158 |
| 5 | Dominater069 | 156 |
| 6 | adamant | 154 |
| 7 | djm03178 | 151 |
| 8 | luogu_official | 150 |
| 9 | awoo | 148 |
| 10 | maomao90 | 145 |
Problem link : http://www.spoj.com/problems/PYTRIP/
I am not able to think of any other solution except brute force and also, i am unable to find any pattern or something. If anyone could please give me a hint in approaching the problem or provide a reference to any material that you think might be helpful would be great.
You can observe that a=sqrt(2*c-1) and b=c-1. Iterate from n to m and check for every i if sqrt(2*i-1)*sqrt(2*i-1) = 2*i-1. For c=1 there is no solution cause b=c-1=0.
C^2=a^2+b^2. So c^2-b^2=(c-b)(c+b)=a^2. You can say that c-b=1 and c+b=a^2. So c-b=1 => b=c-1. Now c+b=a^2 => 2*c-1=a^2. So a = sqrt(2*c-1).
I did for odd c here is full explanation http://codeforces.me/blog/entry/46681
You can use Euclid's formula