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    论可复用的游戏服务器端开发框架(一)

    本文试图以游戏服务器端开发的角度,探讨在需求高度变化的环境下,可重用模块构建的可能性和基本方案。 可复用框架的必要性与可行性 在现代游戏产品的开发中,游戏服务器端程序已经几乎成为了标配。从最简单的正版保护功能,到玩家档案、成就的存储功能,到复杂的主要游戏逻辑运算,游戏服务器端系统都是必不可少的。但是和客户端丰富的游戏引擎不同,服务器端比较少这类可复用的软件产品出现。其原因可能有以下几个:一是欧美、日本的服务器端逻辑一般比较少,所以这类产品的需求也比较少;二是游戏服务器端本身涉及大量不同的运行平台、环境、语

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    第九艺术的降临:游戏如何讲故事

    在我所玩过的游戏中,分为两种,一种是“玩具”类游戏,一种是所谓“演出”类游戏。所谓“玩具”类游戏,主要是通过游戏的玩法来提供乐趣。而“演出”类游戏,则在玩法之上,通过游戏的剧情、美术、音乐等可欣赏的内容,叠加出另外一种乐趣来。 举例来说,《王者荣耀》就是一类“由玩家提供内容”的玩具类游戏,就好像一个足球,让你和其他人一起来玩,好玩与否取决于你和谁玩。说实在的腾讯运营的大多数赚钱的游戏,都是这一类。而《文明》这一类,则是另外一种玩具类游戏,它的游戏内容是游戏自己提供的。这和早期大多数的强调“游戏性”的单机游戏

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    杭电OJ2060-2069

    background: Philip likes to play the QQ game of Snooker when he wants a relax, though he was just a little vegetable-bird. Maybe you hadn’t played that game yet, no matter, I’ll introduce the rule for you first. There are 21 object balls on board, including 15 red balls and 6 color balls: yellow, green, brown, blue, pink, black. The player should use a white main ball to make the object balls roll into the hole, the sum of the ball’s fixed value he made in the hole is the player’s score. The player should firstly made a red ball into the hole, after that he gains red-ball’s value(1 points), then he gets the chance to make a color ball, then alternately. The color ball should be took out until all the red-ball are in the hole. In other word, if there are only color balls left on board, the player should hit the object balls in this order: yellow(2 point), green(3 point), brown(4 point), blue(5 point), pink(6 point), black(7 point), after the ball being hit into the hole, they are not get out of the hole, after no ball left on board, the game ends, the player who has the higher score wins the game. PS: red object balls never get out of the hole. I just illustrate the rules that maybe used, if you want to contact more details, visit http://sports.tom.com/snooker/ after the contest. for example, if there are 12 red balls on board(if there are still red ball left on board, it can be sure that all the color balls must be on board either). So suppose Philp can continuesly hit the ball into the hole, he can get the maximun score is 12 × 1 (12 red-ball in one shoot) + 7 × 12(after hit a red ball, a black ball which was the most valuable ball should be the target) + 2 + 3 + 4 + 5 + 6 + 7(when no red ball left, make all the color ball in hole). Now, your task is to judge whether Philip should make the decision to give up when telling you the condition on board(How many object balls still left not in the hole and the other player’s score). If Philp still gets the chance to win, just print “Yes”, otherwise print “No”.

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