期望DP 一个数只能分解成不大于它的数,那么转移构成拓扑关系. 试了一下预处理出不大于x的质数个数,然而程序并没有变快 /*by SilverN*/ #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<vector> using namespace std; ; int read(){ ,f…

题意 给出一个整数 $N$,每次可以在不超过 $N$ 的素数中等概率随机选择一个 $P$,如果 $P$ 是 $N$ 的约数,则把 $N$ 变成 $N/P$,否则 $N$ 不变.问平均情况下需要多少次随机选择,才能把 $N$ 变成1呢? 分析 本题可以画出一个状态转移图, 例如 $n=6$ 时, $n$ 的每个约数都对应一个状态,每个状态转移都有一定概率,从每个状态出发转移的概率和为1. 设 $f(i)$ 表示当前的数为 $i$ 时接下来需要选择的期望次数,可列出方程: $$f(6) = 1 +…

模拟马尔可夫过程,具体看书. #include<bits/stdc++.h> #define N 1000010 using namespace std; ; void calcprime(){ cnt=;memset(use,,sizeof(use)); ;i<=N;i++){ if(use[i]){prime[++cnt]=i;} ;j<=cnt;j++){ int t=prime[j]*i;if(t>N)break; use[t]=; )break; } } } boo…

[Promise.race] 返回最先完成的promise var p1 = new Promise(function(resolve, reject) { setTimeout(resolve, 500, "one"); }); var p2 = new Promise(function(resolve, reject) { setTimeout(resolve, 100, "two"); }); Promise.race([p1, p2]).then(funct…

golang中的race检测 由于golang中的go是非常方便的,加上函数又非常容易隐藏go. 所以很多时候,当我们写出一个程序的时候,我们并不知道这个程序在并发情况下会不会出现什么问题. 所以在本质上说,goroutine的使用增加了函数的危险系数论go语言中goroutine的使用.比如一个全局变量,如果没有加上锁,我们写一个比较庞大的项目下来,就根本不知道这个变量是不是会引起多个goroutine竞争. 官网的文章Introducing the Go Race Detector给出的例子…

2599: [IOI2011]Race Time Limit: 70 Sec  Memory Limit: 128 MBSubmit: 2590  Solved: 769[Submit][Status][Discuss] Description 给一棵树,每条边有权.求一条简单路径,权值和等于K,且边的数量最小.N <= 200000, K <= 1000000 Input 第一行 两个整数 n, k第二..n行 每行三个整数 表示一条无向边的两端和权值 (注意点的编号从0开始) Output…

Bob’s Race Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads…

题目链接: http://codeforces.com/problemset/problem/214/E Relay Race time limit per test4 secondsmemory limit per test256 megabytes 问题描述 Furik and Rubik take part in a relay race. The race will be set up on a large square with the side of n meters. The gi…

Race Condition(也叫做资源竞争),是多线程编程中比较头疼的问题.特别是Java多线程模型当中,经常会因为多个线程同时访问相同的共享数据,而造成数据的不一致性.为了解决这个问题,通常来说需要加上同步标志“synchronized”,来保证数据的串行访问.但是“synchronized”是个性能杀手,过多的使用会导致性能下降,特别是扩展性下降,使得你的系统不能使用多个CPU资源.  这是我们在性能测试中经常遇见的问题. 可是上个星期我却遇见了相反的情况:因为缺少同步标志也同样会使性能受…

C. The Big Race Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/592/problem/C Description Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of…

Bob’s Race Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4123 Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads…

C - Race to 1 Again Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Description Rimi learned a new thing about integers, which is - any positive integer greater than 1 can be divided by its divisors. So, he is now playing w…

Walking Race   Description flymouse's sister wc is very capable at sports and her favorite event is walking race. Chasing after the championship in an important competition, she comes to a training center to attend a training course. The center has N…

Single, race both get only one emit value from the stream. Single(fn): const source = Rx.Observable.from([1,2,3,4,5]); var single = source.single( x => x === 3); /* (12345)| (source) single( x => x === 3) 3| (single) */ race(...observable): Observab…

UVA 10627 - Infinite Race option=com_onlinejudge&Itemid=8&page=show_problem&category=516&problem=1568&mosmsg=Submission+received+with+ID+13837674" target="_blank" style="">题目链接 题意:一段跑道,A,B分别在两端,速度为u.v,两个人跑到还…

虽说是IOI'95,但是也是挺水的..for 第一问,n最大为50,所以可以直接枚举起点和终点之外的所有点,然后dfs判断是否连通:for 第二问,易知答案一定是第一问的子集,所以从第一问中的答案中枚举,也是用dfs判断. ---------------------------------------------------------------------- #include<cstdio> #include<iostream> #include<cstring>…

数据范围是N:20w, K100w. 点分治, 我们只需考虑经过当前树根的方案. K最大只有100w, 直接开个数组CNT[x]表示与当前树根距离为x的最少边数, 然后就可以对根的子树依次dfs并更新CNT数组和答案. 复杂度是O(NlogN) ------------------------------------------------------------------------------------------ #include<bits/stdc++.h>   using nam…

Bob’s Race Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 378   Accepted: 119 Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his vil…

D. Bicycle Race 题目链接http://codeforces.com/contest/659/problem/D Description Maria participates in a bicycle race. The speedway takes place on the shores of Lake Lucerne, just repeating its contour. As you know, the lake shore consists only of straigh…

Description Maria participates in a bicycle race. The speedway takes place on the shores of Lake Lucerne, just repeating its contour. As you know, the lake shore consists only of straight sections, directed to the north, south, east or west. Let's in…

关于promise我在之前的文章已经应用过好几次,如[js高手之路]Node.js+jade+express+mongodb+mongoose+promise实现todolist,本文就来讲解下promise的常见用法. 为什么会有promise,他的作用是什么? promise主要是为了解决js中多个异步回调难以维护和控制的问题. 什么是promise? 从图中,我们可以看出,Promise是一个函数,这个函数上有在项目中常用的静态方法:all, race, reject,resolve等,原…

题目链接 Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses, and all houses are connected together. To…

Race to 1 Time Limit: 10000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   B Race to 1 Input: Standard Input Output: Standard Output   Dilu have learned a new thing about integers, which is - an…

[BZOJ2599]Race(点分治) 题面 BZOJ权限题,洛谷 题解 好久没写过点分治了... 在ppl的帮助下终于想起来了 orz ppl 首先回忆一下怎么求有没有正好是\(K\)的路径 维护一个表示距离的桶 对于当前重心,依次插入每棵子树的距离值 然后检查是否存在即可 显然加一步,求最短的路径数 那么,把原来的是否存在的01数组 改为记录最短路径数的一个\(int\)数组 每次插入的时候去取\(min\)就行了 #include<iostream> #include<cstdio…

新概念英语(1-69)The car race Which car was the winner in 1995 ? There is  car race near our town every year. In 1995, there was a very big race. There were hundreds of people there. My wife and I were at the race. Our firends Julie and Jack were there too…

一.Pomise.all的使用 Promise.all可以将多个Promise实例包装成一个新的Promise实例.同时,成功和失败的返回值是不同的,成功的时候返回的是一个结果数组,而失败的时候则返回最先被reject失败状态的值. 具体代码如下: let p1 = new Promise((resolve, reject) => { resolve('成功了') }) let p2 = new Promise((resolve, reject) => { resolve('success')…

function mypromise(func){ this.statue = "pending"; this.data = null; this.resolveCallback = []; this.rejectCallback = []; this._final = null; var self = this; var resolve = function(data){ if (data instanceof mypromise) { data.then(resolve,rejec…

Walking Race Time Limit: 10000MS   Memory Limit: 131072K Total Submissions: 4123   Accepted: 1029 Case Time Limit: 3000MS Description flymouse’s sister wc is very capable at sports and her favorite event is walking race. Chasing after the championshi…

一.Pomise.all的使用 常见使用场景 : 多个异步结果合并到一起 Promise.all可以将多个Promise实例包装成一个新的Promise实例.用于将多个Promise实例,包装成一个新的Promise实例. 1.它接受一个数组作为参数. 2.数组可以是Promise对象,也可以是其它值,只有Promise会等待状态改变. 3.当所有的子Promise都完成,该Promise完成,返回值是全部值的数组. 4.如果有任何一个失败,该Promise失败,返回值是第一个失败的子Promi…

public class TestRaceCondition implements Runnable{ public static int counter = 0; public static void main(String[] args) { List<Thread> threads = new ArrayList<>(); //產生threads,加入ArrayList for( int i=0 ; i < 10 ; i++) { TestRaceCondition r…