A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road. Array A contains only 0s and/or 1s: 0 represents a car traveling east, 1 represents a car traveling west. Th

http://codility.com/demo/take-sample-test/beta2010/ 这题以前做的时候是先排序再二分,现在觉得没有必要.首先圆可以看成线段,把线段的进入作为一个事件,出去作为一个事件.注意根据题意,同样的点,进入要在出去之前.那么O(n)扫一遍就可以得到结果.注意的是A[i]+i可能会超出int. // you can also use includes, for example: #include <algorithm> int solution(const

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,

The Game 题意: Description One morning, you wake up and think: "I am such a good programmer. Why not make some money?" So you decide to write a computer game. 一天清晨,你起床后想到:"我是一个这么牛的程序员,为什么不赚点钱呢?"所以,你决定写一个电脑游戏. The game takes place on a re

1.题目大意 A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, these are arithmetic sequence: 1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9 The

Problem C. Painting CottagesTime Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/attachments Description The new cottage settlement is organized near the capital of Flatland. The construction company that is building the settl

链接:http://poj.org/problem?id=1502 MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5249   Accepted: 3237 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed share

题意: 给定n个平面(平面之间相互独立),每个平面上有一些点,并且构成凸集,C和D轮流选一个平面连接两个点画线段,并保证线段之间除了端点之外没有其它交点,当平面上出现一个完整的三角形之后此平面就不能继续画线.最早无法画线的人输.输出赢的人. 解法: 因为n个平面是独立的,所以sg函数满足异或的关系.对于每一个平面,求sg值.对于n个点,连上一条线可以分成 i 和 n-2-i 两个独立的部分.所以该点的子状态为sg[i]^sg[n-i-2](0<=i<=n-2).然后可以计算该点的sg值.打表发

#include<stdio.h> #include<vector> #include<algorithm> using namespace std; struct Node { int x,y; } p; bool cmp(Node a,Node b) { if(a.x<b.x)return true; if(a.x==b.x&&a.y<b.y)return true; return false; } vector<Node>v

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.

https://codility.com/demo/take-sample-test/count_div 此题比较简单,是在O(1)时间里求区间[A,B]里面能被K整除的数字,那么就计算一下就能得到. int solution(int A, int B, int K) { // write your code in C++11 int diff = (B-A+1); int result = diff / K; for (int i = A + result * K; i <= B; i++)

https://codility.com/demo/take-sample-test/min_avg_two_slice 此题要求一个数组子段的最小的平均数(返回第一个数字的index).刚开始想记录sum,也没用,因为所有子段的组合是O(n^2)的.后来看解释发现,最小值必然存在于长度为2或3的子段中(长度为3的子段也无法分解).可以用反证法,如果有一个长度大于3的子段,均值小于之前找出的长度2和3的子段均值,那么必然可以在这个子段中找到均值更小的长度2和3的子段,矛盾. int soluti

今天开始刷刷codility上respectable的题目,难度适中. https://codility.com/demo/take-sample-test/missing_integer 本题是找出数组里第一个非负的整数,要求复杂度O(n).那么首先想到的做法是排序,但这样负责度就上去了.要从nlogn降到n,常见的一个做法是用hashtable,这里就可以用set记录.要注意的是全负的情况,所以这里用了maxVal=0作为初值. #include <unordered_set> using

在写上一随笔之前,在Codility网站上还做了一个道题(非Demo题):CountBoundedSlices,得了60分(害臊呀).今天又重新做了一下这个算法,性能提高了不少,但由于此题不是Demo题,不能重新在Codility网站测试了. 可以从http://codility.com/cert/view/certAWY5VP-D46CA7989XU4XEZT/details看到题目的详细信息. 这里把题目copy出来: An integer K and a non-empty zero-in