[ABC267F] Exactly K Steps

【[ABC267F] Exactly K Steps】的更多相关文章

字典序的第K小数字

今天zyb参加一场面试,面试官听说zyb是ACMer之后立马抛出了一道算法题给zyb:有一个序列,是1到n的一种排列,排列的顺序是字典序小的在前,那么第k个数字是什么?例如n=15,k=7, 排列顺序为1, 10, 11, 12, 13, 14, 15, 2, 3, 4, 5, 6, 7, 8, 9;那么第7个数字就是15.那么,如果你处在zyb的场景下,你能解决这个问题吗题解 https://blog.csdn.net/FJJ543/article/details/81908992 #inc…

第十四届华中科技大学程序设计竞赛 K Walking in the Forest【二分答案/最小化最大值】

链接:https://www.nowcoder.com/acm/contest/106/K 来源:牛客网题目描述 It's universally acknowledged that there're innumerable trees in the campus of HUST. Now you're going to walk through a large forest. There is a path consisting of N stones winding its way to…

[LeetCode] Rotate Array 旋转数组

Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. Note:Try to come up as many solutions as you can, there are at least 3 different ways to solve this pro…

Functional Programming without Lambda - Part 2 Lifting, Functor, Monad

Lifting Now, let's review map from another perspective. map :: (T -> R) -> [T] -> [R] accepts 2 parameters, a function f :: T -> R and a list list :: [T]. [T] is a generic type paramterized by T, it's not the same as T, but definitely shares s…

35.两链表的第一个公共结点[Find the first common node of two linked list]

[题目] 两个单向链表,找出它们的第一个公共结点. 链表的结点定义为: C++ Code 123456 struct ListNode { int m_nKey; ListNode *m_pNext; }; [分析] 这是一道微软的面试题.微软非常喜欢与链表相关的题目,因此在微软的面试题中,链表出现的概率相当高. 如果两个单向链表有公共的结点,也就是说两个链表从某一结点开始,它们的m_pNext都指向同一个结点.但由于是单向链表的结点,每个结点只有一个…

array题目合集

414. Third Maximum Number 给一个非空的整数数组,找到这个数组中第三大的值,如果不存在,那么返回最大的值.要求时间复杂度为o(n) 例如: Example 1: Input: [3, 2, 1] Output: 1 Explanation: The third maximum is 1. Example 2: Input: [1, 2] Output: 2 Explanation: The third maximum does not exist, so the maxi…

189. Rotate Array

Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. 解答:K 如果大于length,对 k 求余 public class Solution { public int[] rotate(int[] nums, int k) { if(nums == null…