x越大越难满足条件,二分,每次贪心的选区间判断是否合法.此题精度要求很高需要用long double,结果要输出分数,那么就枚举一下分母,然后求出分子,在判断一下和原来的数的误差. #include<bits/stdc++.h> using namespace std; typedef long double ld; ; ; struct Seg { int l,r; bool operator < (const Seg& x) const { return l<x.l |…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 二分长度. 显然长度越长.就越不可能. 二分的时候.可以不用管精度. 直接指定一个二分次数的上限就好. 判断长度是否可行.直接用贪心就好. ->贪心(排序区间,尽量让新的区间右端点靠左一点.以便后面的区间有放的地方. 最后得到小数. =>暴力枚举分母i是什么. 然后进行类似一个迭代!?的过程. 如果round(ans*i)/i和ans的差的绝对值更小.则更新分母.分子. [代码] #include <bits/stdc+…

二分找到最大长度,最后输出的时候转化成分数,比较有技巧性. AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cctype> #include <cstring> #include <string> #include <sstream> #include <vector> #include <set> #…

题意:给定 n 个区间,然后把它们变成等长的,并且不相交,问最大长度. 析:首先是二分最大长度,这个地方精度卡的太厉害了,都卡到1e-9了,平时一般的1e-8就行,二分后判断是不是满足不相交,找出最长的.这个题并不难, 就是精度可能控制不好,再就是把小数化成分数时,可能有点麻烦. 代码如下: #include <iostream> #include <cmath> #include <cstdlib> #include <set> #include <…

NEERC2012 A - Addictive Bubbles 题目描述:有一个\(n \times m\)的棋盘,还有不同颜色的棋子若干个,每次可以消去一个同种颜色的联通块,得到的分数为联通块中的棋子数的平方,消掉后高处的棋子会落到下面.设计一个摆法,使得在最优策略下能得到最多的分数. solution 同种颜色的棋子摆在一起即可. 时间复杂度:\(O(nm)\) B - Blind Problem Solving 题目描述:有\(n\)个物品,一个容量为\(c\)的背包,每个物品的大小为\(…

Problem UVA1616-Caravan Robbers Accept: 160  Submit: 1156Time Limit: 3000 mSec Problem Description OnceuponatimetherewasagreedyKing who ordered his chief Architect to build a field for royal cricket inside his park. The Kingwassogreedy, thathewouldno…

Problem UVA1616-Caravan Robbers Accept: 96  Submit: 946Time Limit: 3000 mSec Problem Description Long long ago in a far far away land there were two great cities and The Great Caravan Road between them. Many robber gangs “worked” on that road. By an…

Problem UVA1616-Caravan Robbers Accept: 531  Submit: 2490Time Limit: 3000 mSec Problem Description Input Input will start with a positive integer, N (3 ≤ N ≤ 500) the number of aliens. In next few lines there will be N distinct integers from 1 to N i…

4484: The Graver Robbers' Chronicles 题目连接: http://acm.scu.edu.cn/soj/problem.action?id=4484 Description One day, Kylin Zhang and Wu Xie are trapped in a graveyard. They find an ancient piece of parchment with a cipher string. After discussion and ana…

//: Playground - noun: a place where people can play import UIKit var str = "Hello, playground" enum Movement { case Left case Right case Top case Bottom } let aMovement = Movement.Left switch aMovement { case .Left: print("left") defa…