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本文介绍了一个来自 Brown Sharpie 的博客内容,尽管具体细节未在提供的信息中给出,但该博客通常涉及教育和技术主题。通过提供的图片链接,我们可以推测博客可能讨论了与绘图或教育相关的技术实践。
Frame-independent movement If you run everything we have done so far, you will be able to move the circle, but it won't move uniformly. It will probably be very fast, because currently we have done the movement in a very naive way. Right now your computer will be running the update() function as fast as it can, which means it will probably call it a couple of hundreds of times each second, if not more. If we move the shape by one pixel for every frame, this can count up to several 100 pixels every second, making our little player fly all over the screen. You cannot just change the movement value to something lower, as it will only fix the problem for your computer. If you move to a slower or faster computer, the speed will change again. So how do we solve this? Well, let's look at the problem we are facing. We are having a problem because our movement is frame-dependent. We want to provide the speed in a way that changes depending on the time a frame takes. There is a simple formula you should remember from your old school days. It's the formula that goes: distance = speed * time. Now why is this relevant for us? Because with this formula we can calculate a relevant speed for every frame, so that the circle always travels exactly the distance we want it to travel over one second, no matter what computer we are sitting on. So let's modify the function to what we actually need to make this work. void Game::update(sf::Time deltaTime) { sf::Vector2f movement(0.f, 0.f); if (mIsMovingUp) movement.y -= PlayerSpeed; if (mIsMovingDown) movement.y += PlayerSpeed; if (mIsMovingLeft) movement.x -= PlayerSpeed; if (mIsMovingRight) movement.x += PlayerSpeed; mPlayer.move(movement * deltaTime.asSeconds()); } The major difference we have made here is that we now receive a time value every time we call the update. We calculate the distance we want to travel every frame, depending on how much time has elapsed. We call the time that has elapsed since the last frame delta time (or time step), and often abbreviate it as dt in the code. But how do we get this time? We are lucky because SFML provides the utilities for it.翻译
03-08
Load balancing (负载均衡) refers to efficiently distributing incoming network traffic across a group of backend servers. A load balancing algorithm distributes loads in a specific way. If we can estimate the maximum incoming traffic load, here is an algorithm that works according to the following rule: The incoming traffic load of size S will first be partitioned into two parts, and each part may be again partitioned into two parts, and so on. Only one partition is made at a time. At any time, the size of the smallest load must be strictly greater than half of the size of the largest load. All the sizes are positive integers. This partition process goes on until it is impossible to make any further partition. For example, if S=7, then we can break it into 3+4 first, then continue as 4=2+2. The process stops at requiring three servers, holding loads 3, 2, and 2. Your job is to decide the maximum number of backend servers required by this algorithm. Since such kind of partitions may not be unique, find the best solution -- that is, the difference between the largest and the smallest sizes is minimized. Input Specification: Each input file contains one test case, which gives a positive integer S (2≤N≤200), the size of the incoming traffic load. Output Specification: For each case, print two numbers in a line, namely, M, the maximum number of backend servers required, and D, the minimum of the difference between the largest and the smallest sizes in a partition with M servers. The numbers in a line must be separated by one space, and there must be no extra space at the beginning or the end of the line. Sample Input: 22 Sample Output: 4 1 Hint: There are more than one way to partition the load. For example: 22 = 8 + 14 = 8 + 7 + 7 = 4 + 4 + 7 + 7 or 22 = 10 + 12 = 10 + 6 + 6 = 4 + 6 + 6 + 6 or 22 = 10 + 12 = 10 + 6 + 6 = 5 + 5 + 6 + 6 All requires 4 servers. The last partition has the smallest difference 6−5=1, hence 1 is printed out.请给这一到题设置测试用例,包含最大最小情况,和一些特殊的情况,以及对于回溯算法最好和最坏的情况
06-03
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