codility上面添加了教程.目前只有lesson 1,讲复杂度的……里面有几个题, 目前感觉题库的题简单. tasks: Frog-Jmp: 一只青蛙,要从X跳到Y或者大于等于Y的地方,每次跳的距离为D,问至少跳几次. X,Y,D都是[1..10^9]的整数. 要求时间空间复杂度O(1). 这个题比较简单,就是做除法嘛,我们不知道X是否已经不小于Y了,我加了个判断,不过也就一句话. 代码: // you can also use includes, for example: // #incl…

1.题目: A game for one player is played on a board consisting of N consecutive squares, numbered from 0 to N − 1. There is a number written on each square. A non-empty zero-indexed array A of N integers contains the numbers written on the squares. More…

How to solve this HARD issue 1. Problem: A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A…

首先上题目: A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has an impact factor, which is an integer. Nucleotides of types A,…

https://codility.com/demo/take-sample-test/peaks http://blog.csdn.net/caopengcs/article/details/17491791 其实可以做到O(n) #include <iostream> #include <sstream> using namespace std; int solution(vector<int> &A) { int n = A.size(); int prev…

https://codility.com/programmers/challenges/fluorum2014 http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1273 http://blog.csdn.net/caopengcs/article/details/36872627 http://www.quora.com/How-do-I-determine-the-order-of-visiting-all-leave…

https://codility.com/programmers/challenges/magnesium2014 图形上的DP,先按照路径长度排序,然后依次遍历,状态是使用到当前路径为止的情况:每个节点记录以该节点结束的最长路径,这样加入新的路径时去更新.注意路径是双向的~ #include <vector> #include <algorithm> using namespace std; struct Road { int start; int end; int val; }…

https://codility.com/demo/take-sample-test/max_double_slice_sum 两个最大子段和相拼接,从前和从后都扫一遍.注意其中一段可以为0.还有最后和最前面一个不可能取到~ #include <vector> using namespace std; int solution(vector<int> &A) { vector<int> maxEnd; maxEnd.resize(A.size()); maxEn…

https://codility.com/demo/take-sample-test/fish 一开始习惯性使用单调栈,后来发现一个普通栈就可以了. #include <stack> using namespace std; int solution(vector<int> &A, vector<int> &B) { int size = A.size(); int deadNum = 0; stack<int> stk; // downst…

https://codility.com/programmers/challenges/upsilon2012 求笛卡尔树的高度,可以用单调栈来做. 维持一个单调递减的栈,每次进栈的时候记录下它之后有多少元素,就是以它为根的子树的高度.出栈的时候再更新一次供新进栈者使用. int solution(vector<int> &A) { A.push_back(1000000001); // an element larger than any one in A int size = A.…