[ahu 1248] NBA Finals】的更多相关文章

NBA Finals Time Limit: 1000 ms   Case Time Limit: 1000 ms   Memory Limit: 64 MBTotal Submission: 251   Submission Accepted: 41   Description Consider two teams, Lakers and Celtics, playing a series of NBA Finals until one of the teams wins n games. A…
NBA Finals(0649) Time limit(ms): 1000 Memory limit(kb): 65535 Submission: 404 Accepted: 128   Description   Consider two teams, Lakers and Celtics, playing a series of NBA Finals until one of the teams wins n games. Assume that the probability of Lak…
题目链接:http://acm.swust.edu.cn/problem/649/ Time limit(ms): 1000 Memory limit(kb): 65535 Consider two teams, Lakers and Celtics, playing a series of NBA Finals until one of the teams wins n games. Assume that the probability of Lakers winning a game is…
杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…
转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…
http://acm.hdu.edu.cn/showproblem.php?pid=1248 #include <stdio.h> #include <stdlib.h> #include <string.h> #include <algorithm> #define maxn 11000 using namespace std; int dp[maxn]; int main() { int T, n; scanf("%d", &…
见证历史-- 2013 NBA 热火夺冠之路有感今年NBA季后赛从第一轮看起,到最终的热火夺冠,应该看得是最爽的一次.但一些情节和细节,回忆起来,深有感悟. 1. 做人要低调詹宁斯豪言演黑八雄鹿本赛季常规赛4战热火1胜3负,两队常规赛战绩相差28个胜场,无论阵容配置还是获胜场次都差距明显.雄鹿被看衰在情理之中,但本赛季打热火场均拿23.8分的詹宁斯并不这样认为,他在接受采访时谈到了与热火的首轮对抗,放言雄鹿将上演“黑八”.“我们将用6场比赛取得胜利,”詹宁斯说.这不是詹宁斯第一次口出狂言了,他在…
JavaScript实现简单省市(NBA版)联动 <!DOCTYPE html> <html> <head> <title>JavaScript实现简单省市(NBA版)联动</title> <meta http-equiv="content-type" content="text/html; charset=UTF-8"> <script type="text/javascrip…
原题下载:http://icpc.baylor.edu/download/worldfinals/problems/icpc2013.pdf 题目翻译: 问题描述 有n个机器,每个机器有2个芯片,每个芯片可以放k个电池. 每个芯片能量是k个电池的能量的最小值. 两个芯片的能量之差越小,这个机器就工作的越好. 现在有2nk个电池,已知它们的能量,我们要把它们放在n个机器上的芯片上, 使得所有机器的能量之差的最大值最小. 输入格式 第一行,两个正整数,n和k. 第二行,2nk个整数,表示每个电池的能…
D. Finals in arithmetic 题目连接: http://www.codeforces.com/contest/625/problem/D Description Vitya is studying in the third grade. During the last math lesson all the pupils wrote on arithmetic quiz. Vitya is a clever boy, so he managed to finish all th…