How can we prove this : no of unique outputs to (i*j)%k where i can be any number j and k are known is k/__gcd(k,j);
# | 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 | 160 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | Dominater069 | 154 |
8 | awoo | 154 |
10 | luogu_official | 151 |
How can we prove this : no of unique outputs to (i*j)%k where i can be any number j and k are known is k/__gcd(k,j);
Name |
---|
You need to find for which a there exists a solution to ij ≡ a (mod k) where i is a variable. Obviously if (k, j) = 1 then j has an inverse mod j so for any a < k there exists an unique solution.
Now suppose (k, j) = d! = 1, then your equation becomes equivalent to finding an l such that ij = lk + a or ij + ( - l)k = a. A pair i, l exists if and only if , so a = dx where x has some random value. Now you have d * x < k so x < k / d, so x can range from 0 to k / d - 1, so there are exactly k / d variables for which there exists a solution.