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Codeforces Round 952(DIV 4) doubt, Problem G
For numbers with l+1 digits, there are floor(9/k) (say a) options per digit, hence the total number of numbers should be a^(l+1). Similarly, for l+2 digits, a^(l+2). So on for r digits, a^(r).
Shouldn't the answer be the sum of these numbers?
a^(l+1) + a^(l+2) + .... + a^r
| Rev. | Lang. | By | When | Δ | Comment | |
|---|---|---|---|---|---|---|
| en2 |
|
ayushanshul07 | 2024-06-15 07:04:55 | 4 | Tiny change: 'loor(9/k) (say a) o' -> 'loor(9/k) + 1 (say a) o' | |
| en1 |
|
ayushanshul07 | 2024-06-15 06:53:32 | 347 | Initial revision (published) |
