Блог пользователя mistborn

Автор mistborn, 5 лет назад, По-английски

Say we are given the number of occurrences of each letter (lower case english alphabet). We have to find the number of all k length strings (unique) that can be formed using these letters (MOD 1e9 +7).

Example: a:2, b:1, c:1

k = 2

aa ab ac bc ca cb ba

Answer: 7

Constraints:

k <= 1e5

Total number of characters <= 1e5

I couldn't solve it. Any input is appreciated.

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5 лет назад, # |
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Right now I can only think of an ugly solution involves FFT.

The answer is the coefficient of $$$x^{k}$$$ in the result of $$$\prod_{i=1}^{n}(\sum_{j=0}^{min(a_i,k)}\frac{x^j}{j!})$$$ times $$$k!$$$ if $$$a_i$$$ denotes the max number of ith character we may have in this string.

Multiplication of two polynomials can be done in time $$$O(|degree \ of \ polynomial|\times log)$$$. So if we carefully multiply the two polynomials with the smallest degrees each time, this solution should be able to squeeze in the time limit.