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Leetcode2829.k-avoiding数组的最小总和给你两个整数n和k。对于一个由不同正整数组成的数组,如果其中不存在任何求和等于k的不同元素对,则称其为k-avoiding数组。返回长度为n的k-avoiding数组的可能的最小总和。n个不同正整数的最小总和,那就是从1
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InRust,referencecyclesoccurwhentwoormoreobjectsmutuallyreferenceeachother,formingacircularchain.Inthissituation,thereferencecountbetweenobjectsneverbecomeszero,leadingtomemoryleaksandresourceleaks.Thisblogpostwilldi
- 2023-10-18Go - Avoiding Test Fixtures in Performance Tests
Problem: Youwanttocustomizetheperformanceteststoavoidbenchmarkingtestfixtures.Solution: Youcanstart,stop,andresetthebenchmarktimersusingtheStartTimer,StopTimer,andResetTimer,respectively.Thiswillallowyoutheflexibilityt
- 2023-10-13Codeforces Global Round 11 A. Avoiding Zero
给一个大小为\(n\)的数组\(a_1,a_2,\cdots,a_n\)。你需要构造一个大小为\(n\)的数组\(b\)且满足以下条件:数组\(b\)是数组\(a\)的冲排列对于\(\forallk=1,2,\cdots,n\),\(\sum_{i=1}^{k}b_i\neq0\)。输出任意一组构造,或者回答不可能。若\(\sum_{i