| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 162 |
| 2 | adamant | 148 |
| 3 | Um_nik | 145 |
| 4 | Dominater069 | 143 |
| 5 | errorgorn | 141 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 135 |
| 7 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 132 |
|
0
in problem E where there is equation for b2-b1 we should substract (n-1)*a2 instead of n*a2 |
|
0
Solve problems |
|
0
Gratz |
|
0
0 is a special case. When you have more than (n+1)/2 zeros then MEX != 0; If 0 isn't MEX we look at other numbers from 1 to n. If n-number_of_zeros_in_our_array>=1 it means that this number is MEX, because you can just make an array like that: {actual_number,x,x,x,0,0,0}. Zeros are ignored so we put them at the end to not damage our result. x- are all numbers excluding zeros. |
|
0
Could D be solved this way but counting decreasing values instead of increasing? |
|
+1
you could just copy pi digits from the sample input xD |
| Name |
|---|


