ExponentiationTime Limit: 500MS Memory Limit: 10000KTotal Submissions: 155886 Accepted: 37967 Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, th…
PKU OJ 1002 487-3279 487-3279 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialin…
这个估计是里面第二简单的了,因为第一简单的是求a+b 哈哈,一submit就ac了 题目如下: Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to…
http://bailian.openjudge.cn/tm2018/G/ #include <iostream> #include <vector> #include <string> using namespace std; const int N = 2001; bool flag = false; //父亲节点 int Father[N]; // 要想与根节点之间的关系如何表示?可以用0表示与根节点同性别, //用1表示与根节点是不同性别,可以用%2,来判断两只…
Problem Link: https://oj.leetcode.com/problems/validate-binary-search-tree/ We inorder-traverse the tree, and for each node we check if current_node.val > prev_node.val. The code is as follows. # Definition for a binary tree node # class TreeNode: #…
Problem Link: https://oj.leetcode.com/problems/recover-binary-search-tree/ We know that the inorder traversal of a binary search tree should be a sorted array. Therefore, we can compare each node with its previous node in the inorder to find the two…
Problem Link: https://oj.leetcode.com/problems/same-tree/ The following recursive version is accepted but the iterative one is not accepted... # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left =…
Problem Link: https://oj.leetcode.com/problems/symmetric-tree/ To solve the problem, we can traverse the tree level by level. For each level, we construct an array of values of the length 2^depth, and check if this array is symmetric. The tree is sym…
Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree level by level using BFS method. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # se…
