题目链接: http://codeforces.com/problemset/problem/155/E E. Double Profiles time limit per test 3 secondsmemory limit per test 256 megabytes 问题描述 You have been offered a job in a company developing a large social network. Your first task is connected wit…

Double Profiles Time Limit: 3000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 154C64-bit integer IO format: %I64d      Java class name: (Any)   You have been offered a job in a company developing a large social net…

You have been offered a job in a company developing a large social network. Your first task is connected with searching profiles that most probably belong to the same user. The social network contains n registered profiles, numbered from 1 to n. Some…

A. Search for Pretty Integers [题目链接]:http://codeforces.com/contest/872/problem/A time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given two lists of non-zero digits. Let's call an in…

STL 中 sort 函数用法简介 做 ACM 题的时候,排序是一种经常要用到的操作.如果每次都自己写个冒泡之类的 O(n^2) 排序,不但程序容易超时,而且浪费宝贵的比赛时间,还很有可能写错. STL 里面有个 sort 函数,可以直接对数组排序,复杂度为 n*log2(n) . 使用这个函数,需要包含头文件#include <algorithm>. 这个函数可以传两个参数或三个参数.第一个参数是要排序的区间首地址,第二个参数是区间尾地址的下一地址.也就是说,排序的区间是 [a,b) .简单…

题目链接:http://codeforces.com/problemset/problem/70/D 本题关键:在log(n)的复杂度内判断点在凸包 或 把点插入凸包 判断:平衡树log(n)内选出点所属于的区域 插入:平衡树log(n)内选出点所属于的区域, 与做一般凸包的时候类似,分别以该点向左右两边进行维护, 一直删除不满足凸包的点,直到所有点满足凸包为止. 水平序: 可以用2个平衡树分别维护上下2个半凸包,具体实现时可以把其中一个半凸包按y轴对称以后,那么2个半凸包的维护就是同一种方法,…

DataTable  查询出某个字段为varchar 类型的.不过里面存的为数字,需要进行排序.可是如果直接排序就会不对.因为为varchar类型的,需要转换一下. 方法一: dt.Columns.Add("iPersonOrderSort", System.Type.GetType("System.Double")); dt.Columns["iPersonOrderSort"].Expression = "Convert(tagva…

当values都是整形时,按照Hash的Values排序: h = {'a'=>1,'b'=>2,'c'=>5,'d'=>4} h.sort {|a,b| a[1]<=>b[1]} 输出:[["a", 1], ["b", 2], ["d", 4], ["c", 5]] 当我们需要对Hash进行排序时,不能像数组那样简单的使用sort方法,因为数组中的数据类型都是一样的(整型),Hash中的数…

题目链接: http://codeforces.com/problemset/problem/484/B 题意: 求a[i]%a[j] (a[i]>a[j])的余数的最大值 分析: 要求余数的最大值非常明显a[i]越接近a[j]的倍数则余数越大 ,因此我们将全部的元素从大到小排序 : 然后枚举a[j]的倍数 ,二分查找小于a[i]倍数的最大值,然后更新余数的最大值. 代码例如以下: #include <iostream> #include <cstdio> #include…

You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected. You should assign labels to all vertices in such a way that: Labels form a valid pe…