摘要: Let G be an additive finite abelian p-group. For a given (long) sequence S in G and some element g ∈ G, we investigate the number of subsequences of S which have sum g. This refines some classical results of J.E. Olson and recent results of I. Koutis....
W. D. Gao. On the number of subsequences with given sum. Discrete Math., 195:127-138, 1999.Gao, W., 1999. On the number of subsequences with given sum. Discrete Math. 195 (1-3), 127-138.On the number of subsequences with given sum, Discrete Math. 195 (1999), 127 - 138....
Given an array of integersnumsand an integertarget. Return the number of non-empty subsequences ofnumssuch that the sum of the minimum and maximum element on it is less or equal totarget. Since the answer may be too large, return it modulo109 + 7. Example 1: Input: nums = [3,5,6,...
Given an array of integers nums and an integer target. Return the number of non-empty subsequences of nums such that the sum of the minimum and maximum element on it is less or equal to target. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: nums = ...
Algebraic aspects of increasing subsequences These include: integral formulae for the distribution of the longest increasing subsequence of a random involution with constrained number of fixed points; ... EM Rains,J Baik - 《Duke Mathematical Journal》 被引量: 306发表: 2001年 On Weighted Sequence ...
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The reason is that for each character of the string, we have two options: to pick or leave it. On the other hand, the space complexity of this algorithm is , where is the sum of the length of all possible subsequences of the given string . The reason behind this complexity is that ...
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5. Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int. ...
Letsmagic(sum,cur)magic(sum,cur)is the number of valid subsequences whose selected sum issumsumand current selecting element iscurcur curcuris init as11(smallest element) and recursive stop when it greater thannn(largest element) sumsumis in range[0,n][0,n]when it equalnnthen we found ...