Could someone give me a hint for this problem?
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
| 8 | ksun48 | 3461 |
| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 146 |
| 3 | Proof_by_QED | 146 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 141 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
Could someone give me a hint for this problem?
| Name |
|---|



In this problem we must use bfs on a graph, where each vertex is the remainder of the division by n.
Why do you minusing it? Can't solve? That's nice problem
Oh, i thought that there is some condition like length of answer must be less than 1000. Of course, this problem is obvious: number 1111111....1111000000...000, with 9φ(n) of "1" and 30 "0" is an answer
Would you explain more?
Check out Euler Function and Euler`s theorem.
We can reduce our problem to n so that gcd(n, 2) = gcd(n, 5) = 1. Now, we know that 10φ(n) - 1 is divisible on n, i.e. 99999...999 with φ(n) "9" is divisible on n. It`s easy to check that 1111111....1111000000...000, with 9φ(n) of "1" and 30 "0" is divisible on 10φ(n) - 1 so we are done.
I'm not sure if that always generates the smallest multiple.
Wow, really:( I didn`t see it:( But it's obvious anyway