1044 Shopping in Mars (25 分) Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and som…

1044 Shopping in Mars(25 分) Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some…

1044 Shopping in Mars (25 分)   Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and s…

https://pintia.cn/problem-sets/994805342720868352/problems/994805439202443264 Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

题目 Shopping in Mars is quite a diferent experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

分析: 考察二分,简单模拟会超时,优化后时间正好,但二分速度快些,注意以下几点: (1):如果一个序列D1 ... Dn,如果我们计算Di到Dj的和, 那么我们可以计算D1到Dj的和sum1,D1到Di的和sum2, 然后结果就是sum1 - sum2: (2): 那么我们二分则要搜索的就是m + sum[i]的值. #include <iostream> #include <stdio.h> #include <algorithm> #include <cstr…

双指针. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<algorithm> using namespace std; +; long long a[maxn]; in…

n,m然后给出n个数让你求所有存在的区间[l,r],使得a[l]~a[r]的和为m并且按l的大小顺序输出对应区间.如果不存在和为m的区间段,则输出a[l]~a[r]-m最小的区间段方案. 如果两层for循环l和r的话,会超时,实际上for循环一遍即可. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <vector> #…

#include <cstdio> #include <cstdlib> #include <vector> #include <climits> #include <algorithm> using namespace std; int main() { int N, M; scanf("%d%d", &N, &M); vector<int> nums(N); vector<pair<…

题意: 输入一个正整数N和M(N<=1e5,M<=1e8),接下来输入N个正整数(<=1e3),按照升序输出"i-j",i~j的和等于M或者是最小的大于M的数段. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; vector<pair<int,int> >ans; int main()…

1044. Shopping in Mars (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken o…

一.技术总结 可以开始把每个数都直接相加当前这个位置的存放所有数之前相加的结果,这样就是递增的了,把i,j位置数相减就是他们之间数的和. 需要写一个函数用于查找之间的值,如果有就放返回大于等于这个数的右下标,函数中采用引用的传参方式,就不需要返回值了. 在处理存放下标问题上,如果发现有更小的数时,直接清空之前存放的值. 二.参考代码 #include<algorithm> #include<iostream> #include<vector> using namespa…

1044 Shopping in Mars(25 分) Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1041. Be Unique (20) 题意 给出 N (<=105)个数(数值范围为 [1, 104]),找到其中不重复的第一个数字.比如给出5 31 5 88 67 88 17 , 答案是 31 . 分析 简单模拟题,开一个大数组int a[10001];,以读入的数为下标,记录 count:a[index] ++;.结果输出第一个存储为 1 的下标:if (a[index] == 1). 1042. Shuffling Machine…

导语 昨天上午,微信在广州举办了微信公开课Pro.于是,精神哥这两天的朋友圈被小龙的"八不做"刷屏了.小伙伴们可能不知道,下午,微信公开课专门开设了技术分论坛.在分论坛中,微信开源了跨平台的网络组件Mars. 之前Bugly也为大家介绍过Mars的一些相关模块的内容: 微信终端跨平台组件 mars 系列(一) - 高性能日志模块xlog 微信终端跨平台组件 mars 系列(二) - 信令传输超时设计 今天,微信的小伙伴和大家一起聊聊Mars的设计原则以及发展史. 背景 2012 年中,…

本文来自于腾讯Bugly公众号(weixinBugly),未经作者同意,请勿转载,原文地址:http://mp.weixin.qq.com/s/9DJxipJaaBC8yC-buHgnTQ 作者简介: 曹少琨,微信移动客户端开发工程师,2014年加入腾讯,从事手Q跨平台移动服务框架 MSF 的开发及优化工作,2015 年转入微信终端基础开发组参与网络组件的开发及维护工作. 前言 mars 是微信官方使用 C++ 编写的业务性无关.平台性无关的终端基础组件,目前在微信 Android.iOS.Wi…

1.前言 关于微信内部正在使用的网络层封装库Mars开源的消息,1个多月前就已满天飞(参见<微信Mars:微信内部正在使用的网络层封装库,即将开源>),不过微信团队没有失约,微信Mars 于2016年12月28日正式公开源码(源码地址:https://github.com/Tencent/mars,也可从本文文末的附件下载之,Android版演示程序可以从文末的附件中下载). 之前无论是微信团队还是手机QQ团队,都以腾讯公司的名义在Github开源了数个工程,但这些工程所受的关注度远不及Mar…

题意: 求小于n (1 ≤ n ≤ 10^8)的数中,与n互质的数的四次方和. 知识点: 差分: 一阶差分: 设  则    为一阶差分. 二阶差分: n阶差分:     且可推出    性质: 1. 2. 差分序列: 给你一列数 a[i][1],a[i][2],a[i][3],a[i][4],a[i][5]…… 那么a[i][j]=a[i-1][j+1]-a[i-1][j], 即后一行是上一行相邻两项的差(第一行除外). 如果给你一个多项式, 比如 f(x)=(x+1)*(x+2)*……*(x…

本文来自于腾讯bugly开发者社区,未经作者同意,请勿转载,原文地址:http://dev.qq.com/topic/581c2c46bef1702a2db3ae53 Dev Club 是一个交流移动开发技术,结交朋友,扩展人脉的社群,成员都是经过审核的移动开发工程师.每周都会举行嘉宾分享,话题讨论等活动. 本期,我们邀请了 腾讯 WXG Android 高级工程师"闫国跃",为大家分享<微信mars 的高性能日志模块 xlog>. 大家好 我是来自腾讯微信的闫国跃,很荣幸…

来到机房刷了一道水(bian’tai)题.题目思想非常简单易懂(我的做法实际上参考了Evensgn 范学长,在此多谢范学长了) 题目摆上: 1044: [HAOI2008]木棍分割 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 3162  Solved: 1182[Submit][Status][Discuss] Description 有n根木棍, 第i根木棍的长度为Li,n根木棍依次连结了一起, 总共有n-1个连接处. 现在允许你最多砍断m个…

近日将自己的Ubuntu从14.04LTS升级到了16.04LTS,顿时发现Eclipse不能正常工作了,到Ubuntu的官网上转了一圈发现以下解决方案: 症状: [1]Eclipse启动很慢; [2]Eclipse菜单无法弹出 [3]Eclipse运行的超级慢 Eclipse版本: Mars Ubuntu版本 16.04LTS 解决方法: 1) 在Console中,更改环境变量SWT_GTK3 export SWT_GTK3= 2) 尝试在Console中用命令行的方式启动Eclipse.注意…