[problem:1985G]↵
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For numbers with l+1 digits, there are floor(9/k) + 1 (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).↵
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Shouldn't the answer be the sum of these numbers?↵
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a^(l+1) + a^(l+2) + .... + a^r↵
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↵
For numbers with l+1 digits, there are floor(9/k) + 1 (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↵
↵
