D. Fox And Jumping 题目连接: http://codeforces.com/contest/510/problem/D Description Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at…

Fox And Jumping 题目链接:http://codeforces.com/problemset/problem/512/B dp 若所选卡片能到达区间内任意点,那么所选卡片的最大公约数为1(a*x+b*y=gcd(a,b)=1). 定义状态dp[i]:获得i需要的最小的代价. 代码如下: #include<cstdio> #include<map> #include<iostream> #define LL long long using namespace…

B. Fox And Jumping time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negativ…

Codeforces题号:#510D 出处: Codeforces 主要算法:map+DP 难度:4.6 思路分析: 题意:给出n张卡片,分别有l[i]和c[i].在一条无限长的纸带上,你可以选择花c[i]的钱来购买卡片i,从此以后可以向左或向右条l[i]个单位.购买其他卡片后,可以获得更多的跳跃单位.先要求至少花多少元钱才能够任意跳到纸带上任意一个位置.若不行,输出-1. 首先分析如果只有两个技能的情况.若这两个技能的跳跃长度有最大公约数(x),且满足(x > 1),则一定能跳到任意一个位置.…

根据裴蜀定理,当且仅当选出来的集合的L[i]的gcd等于1时,才能表示任何数. 考虑普通的dp,dp[i][j]表示前i个数gcd为j的最少花费,j比较大,但状态数不多,拿个map转移就好了. $ \bigodot $ 技巧&套路: 裴蜀定理,gcd为1表示任何数. 当状态数不多的时候,map暴力转移dp. #include <bits/stdc++.h> using namespace std; void Work(int &x) { ; i * i <= x; ++i…

[题目链接]:http://codeforces.com/contest/510/problem/D [题意] 你可以买n种卡片; 每种卡片的花费对应c[i]; 当你拥有了第i种卡片之后; 你可以在任意位置x跳到x-l[i]或x+r[i]; 问你至少要花费多少钱买卡片,使得你能够跳跃到坐标轴上的任意一个整数点; [题解] 有个结论: 直接记下来吧 如果gcd(a,b)==1,那么所有的点就都能跳跃到了; 所以问题就转化为,给你n个长度,让你在这n个长度中选取若干个; 使得它们的gcd为1; 且要…

题目链接 题意说的是,有n种卡片,使用第i种卡片可以使当前自己在数轴上的位置移动 l[i],要获得使用第i种卡片的代价是 c[i],求能使自己移动到数轴上任意位置的最小代价,如果不可能则输出-1 当前所拥有的卡片由1->n,逐渐调整map里的值 #include<bits/stdc++.h> using namespace std; ; int n; int c[N],l[N]; map<int,int> dp; int gcd(int a,int b) { return b…

A. Fox And Snake 代码可能有点挫,但能够快速A掉就够了. #include <cstdio> int main() { //freopen("in.txt", "r", stdin); int n, m; scanf("%d%d", &n, &m); ; i <= n; ++i) { == || i % == ) ; j < m; ++j) printf("#"); ==…

D - Fox And Jumping Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 512B Description Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integer…

Fox And Jumping CodeForces - 510D Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0. There are also n cards, each card…

CF习题集三 一.CF8C Looking for Order 题目描述 \(Lena\)喜欢秩序井然的生活.一天,她要去上大学了.突然,她发现整个房间乱糟糟的--她的手提包里的物品都散落在了地上.她想把所有的物品都放回她的手提包.但是,这里有一点问题:她一次最多只能拿两个物品,她也不能移动她的手提包.并且,因为她爱整洁的习惯,如果她拿起了一个物品,她也不能将它放在其他地方,除非放回她的手提包. \(Lena\)把她的房间划分为了一个平面直角坐标系.现在Lena给你她的手提包和每个散落的物品的坐…

/* Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two players.…

Bungee JumpingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1205 Accepted Submission(s): 528 Problem DescriptionOnce again, James Bond is fleeing from some evil people who want to see him dead. F…

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two players. It consi…

Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two…

Cryptolocker是臭名昭著的勒索程序,使用AES加密后密钥回传,用户除了缴纳赎金之外基本无法解密数据. 近日,知名安全公司Fire-Eye与Fox IT联合推出了针对该勒索程序的解锁网站 https://www.decryptcryptolocker.com/ ©…

Webus Fox是基于网页的在线教学.视频会议软件,不用安装,直接使用.它提供文本.语音.视频聊天,文件共享.电子白板等功能. 1. 登录 访问 http://flash.webus.cn/#,用自己的名字登录: 可以同时登录二个不同的帐号体验后面的功能 登录后,点击允许语音.视频: 2. 文件共享.电子白板 2.1 上传PDF文件 点击“上传”按钮,上传一个PDF文件(目前文件的名必须是英文的),上传完后,会出现在“文件分享”列表里 2.2 分享文件(电子白板) 右健你想要演示的文件: 点击…

B. Fox Dividing Cheese time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Two little greedy bears have found two pieces of cheese in the forest of weight a and b grams, correspondingly. The be…

题目描述 A pangram is a phrase that includes at least one occurrence of each of the 26 letters, ‘a’. . .‘z’. You’re probably familiar with this one: “The quick brown fox jumps over the lazy dog.”Your job is to recognize pangrams. For phrases that don’t c…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34052    Accepted Submission(s): 15437 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

E. Super Jumping! Jumping! Jumping! Time Limit: 1000ms Memory Limit: 32768KB 64-bit integer IO format:      Java class name: Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, an…

上次在<在线教学.视频会议软件 Webus Fox(1)文本.语音.视频聊天及电子白板基本用法>里介绍了软件的基本用法.本文主要介绍服务器端如何配置.开发. 1. 配置 1.1 IIS配置 Fox支持最基本的.net Framework4.0和IIS6/IIS7. 在IIS7中,对应应用程序池,需要配置为经典模式,支持.net4.0 1.2 web.config配置 Fox服务器端是暂时是host在IIS上,将来将Host在Windows Service上.对于IIS的配置,web.confi…

本文主要介绍webus fox 客户端的配置及接口说明. 1. 文件列表和配置 1.1 文件列表 1.2 common.xml 配置 根据服务器端的部署, 替换[ServerUrl] , [RtmpPort], [GatewayPort] [GatewayPort]: 指的是iis 里的site的port 2. 相关接口说明 2.1. 消息 2.1.1 MessageClient/MsgClient.as API名称 start 应用场景 启动message消息, 连接服务并登陆 异常   AP…

题意:给一个点(x,y),给一些步长delta1,delta2...deltaN,问从(0,0)严格按照步长走完N步后能否正好到达(x,y)点. 解法:其实就是判断这些线段和(0,0)-(x,y)这条线段能否构成一个多边(角?)形的问题,只需判断最长的边是否不大于于所有边长和的一半即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #inclu…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32564    Accepted Submission(s): 14692 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

Description Farmer John's cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump. The local witch doctor has mixed up P (1 <= P <= 150,000) potions to aid the cows in their quest t…

HDU 1087 题目大意:给定一个序列,只能走比当前位置大的位置,不可回头,求能得到的和的最大值.(其实就是求最大上升(可不连续)子序列和) 解题思路:可以定义状态dp[i]表示以a[i]为结尾的上升子序列的和的最大值,那么便可以得到状态转移方程 dp[i] = max(dp[i], dp[j]+a[i]), 其中a[j]<a[i]且j<i; 另外每个dp[i]可以先初始化为a[i] 理解:以a[i]为结尾的上升子序列可以由前面比a[i]小的某个序列加上a[i]来取得,故此有dp[j]+a[…

B. Fox Dividing Cheese time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Two little greedy bears have found two pieces of cheese in the forest of weight a and b grams, correspondingly. The be…

http://acm.hdu.edu.cn/showproblem.php?pid=1155 Bungee Jumping Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 811    Accepted Submission(s): 349 Problem Description Once again, James Bond is fle…

C. Fox And Names 题目连接: http://codeforces.com/contest/510/problem/C Description Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in…