题目链接:problemId=5376">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376 Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative ch…

Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboar…

Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboar…

Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboar…

题目链接 参考博客:http://blog.csdn.net/napoleon_acm/article/details/40020297 题意:给定n*m的空棋盘 每一次在上面选择一个空的位置放置一枚棋子,直至每一行每一列都至少有一个棋子,求放置次数的期望 分析: dp[i][j][k] 表示当前用了<=k个chess ,覆盖了i行j列(i*j的格子 每行至少一个,每列至少一个)的概率. dp[i][j][k] 由 dp[i][j][k-1] , dp[i-1][j][k-1], dp[i][j…

Description Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns. Every day after work, Edward will place…

一个n行m列的棋盘,每天可以放一个棋子,问要使得棋盘的每行每列都至少有一个棋子 需要的放棋子天数的期望. dp[i][j][k]表示用了k天棋子共能占领棋盘的i行j列的概率. 他的放置策略是,每放一次,就会有四种可能 1)增加一行一列 2)增加一行 3)增加一列 4)不变 所以他放置的概率就可以求出来,每次放下的概率就是当前能放的点除以总的空的点数. 最后统计期望的时候需要统计在第k天刚好符合占满n行m列的概率,就是dp[i][j][k]-dp[i][j][k-1] #include <cstd…

ZOJ Problem Set - 3822 Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a lar…

Domination Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3822 Description Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends.…

Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboar…

Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns. Every day after work, Edward will place a chess piec…

题目链接:zoj 3822 Domination 题目大意:给定一个N∗M的棋盘,每次任选一个位置放置一枚棋子,直到每行每列上都至少有一枚棋子,问放置棋子个数的期望. 解题思路:大白书上概率那一张有一道类似的题目,可是由于时间比較久了,还是略微想了一下. dp[i][j][k]表示i行j列上均有至少一枚棋子,而且消耗k步的概率(k≤i∗j),由于放置在i+1~n上等价与放在i+1行上,同理列也是如此.所以有转移方程: dp[i][j][k+1]+=dp[i][j][k]∗(n−k)(S−k) d…

那天在机房做的同步赛,比现场赛要慢了一小时开始,直播那边已经可以看到榜了,所以上来就知道A和I是水题,当时机房电脑出了点问题,就慢了好几分钟,12分钟才A掉第一题... A.Average Score 题目大意:给定A序列和B序列,长度分别是n和m,告诉你A序列中的n-1个数和B序列的m个数,求剩下的那个A序列中的数满足:将这个数从A序列移除,然后添加到B序列,使得A序列的平均值变小,B序列的平均值变大.求这个数的取值范围(是整数) 解题思路:求出A序列剩下的n-1个数的平均值,和B序列的平均值…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=5381 Information Theory is one of the most popular courses in Marjar University. In this course, there is an important chapter about information entropy. Entropy is the average amount o…

下午就要坐卧铺赶回北京了.闲来无事.写个总结,给以后的自己看. 因为孔神要保研面试,所以仅仅有我们队里三个人上路. 我们是周五坐的十二点出发的卧铺,一路上不算无聊.恰巧邻床是北航的神犇.于是下午和北航神犇玩了一段时间的杀人游戏,晚上还旁观昂神与众神谈论职场形式,未来出路,认为听着有豁然开朗的感觉(听说网上卖菜比搞IT赚的很多其它,也是醉了).无限YM中...只是尽管是卧铺,想入睡也不是那么easy.睡睡醒醒,也挺折腾 第二天七点多到的牡丹江.感觉尽管都是东北,只是这个小城显然不如我家沈阳繁华,尤…

不知道怎样说起-- 感觉还没那个比赛的感觉呢?如今就结束了. 9号.10号的时候学校还评比国奖.励志奖啥的,由于要来比赛,所以那些事情队友的国奖不能答辩.自己的励志奖班里乱搞要投票,自己又不在,真是无语了--烦得要死.然后在这些事情还没处理好之前我们就这样10号中午从地大去北京站上火车了--那时真感觉这场带着这样的心情来现场赛感觉要打铁了-- 然后10号晚上队友的国奖让琦神帮答辩完了.得国奖无疑了.然后自己的励志奖也定下来一定得了.在火车上的我们也松了一口气.不能由于来比赛国奖励志奖都不得是不-…

///dp[i][j][k]代表i行j列件,并把一k的概率 ///dp[i][j][k]一种常见的方法有四种传输 ///1:dp[i-1][j][k-1] 可能 (n-(i-1))*j/(n*m-(k-1)) ///2:dp[i][j-1][k-1] 概率为 i*(m-(j-1))/(n*m-(k-1)) ///3:dp[i-1][j-1][k-1] 概率为 (n-(i-1))*(m-(j-1))/(n*m-(k-1)) ///4:dp[i][j][k-1] 概率为 (i*j-(k-1))/(n…

3799567 2014-10-14 10:13:59                                                                     Accepted                                                             3822 C++ 1870 71760 njczy2010 3799566 2014-10-14 10:13:25                            …

E - Domination Time Limit:8000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu Submit Status Description Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he…

题意: 给N×M的棋盘.每天随机找一个没放过棋子的格子放一个棋子 问使得每一个每列都有棋子的天数期望 思路: dp[i][j][k] 代表放了i个棋子占了j行k列 到达目标状态的期望 然后从 dp[n*m][n][m] 往后递推就好了. 由于知道了有i个棋子 比如一个状态dp[6][3][3] x x x o o o x o o o o o x o x o o o o o o o o o 对于 dp[i+1][3][3] 事实上就是3*3剩下的空再放一个,概率就是(j*k-i) / (n*m-i…

ZOJ 2819 Average Score Time Limit: 2 Sec  Memory Limit: 60 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373 Description Bob is a freshman in Marjar University. He is clever and diligent. However, he is not good at math, especia…

首先赞一下题目, 好题 题意: Marjar University has decided to upgrade the infrastructure of school intranet by using fiber-optic technology. There are N buildings in the school. Each building will be installed with one router. These routers are connected by optic…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5374 思路:题目的意思是求树上的两点,使得树上其余的点到其中一个点的最长距离最小.可以想到这题与树直径有关,我们可以这样做,首先求出树的直径,然后取出树的中点以及与该中点相邻,并且是直径上的一个点,这样就把这棵树划分为两颗子树,然后分别求出这两棵树的直径,最后要选择的两个点分别就是这两棵树的直径上的中点. 一开始是用dfs写的,结果爆栈了,改成bfs就过了. #in…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373 题目意思: 有两个class:A 和 B,Bob 在 Class A 里面.现在给出 Class A(n-1人) 和 Class B(m人) 所有人的分数,除了Bob,所以Class A 少了一个人.现在需要找出 Bob 最大可能的分数和最少可能的分数,使得他在Class A 里面拉低平均分,而在Class B 里面提高平均分. 由于数据量不大,所以可以暴力枚…

I - Information Entropy Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Description Information Theory is one of the most popular courses in Marjar University. In this course, there is an important chapter abo…

套公式 Sample Input 33 bit25 25 50 //百分数7 nat1 2 4 8 16 32 3710 dit10 10 10 10 10 10 10 10 10 10Sample Output 1.5000000000001.4808108324651.000000000000 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm>…

题意:给出A班和B班的学生成绩,如果bob(A班的)在B班的话,两个班级的平均分都会涨.求bob成绩可能的最大,最小值. A班成绩平均值(不含BOB)>A班成绩平均值(含BOB) && B班成绩平均值(不含BOB)< B班成绩平均值(含BOB) 化简后得 B班成绩平均值(不含BOB) < X < A班成绩平均值(不含BOB) Sample Input 24 35 5 54 4 36 55 5 4 5 31 3 2 2 1Sample Output 4 42 4 #…

题目链接:problemId=5383">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5383 Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since ever…

题意: 给你三个均匀k面筛子. 分别有k1 k2 k3个面,每个面朝上的概率是相等的. 如果第一个筛子出现a第二个筛子出现b第三个筛子出现c那么置零. 否则在当前和加上三个点数之和. 求当前和大于n需要的步数的期望. 思路: 一开始状态转移搞错了,手推公式交了WA,后来想了想状态转移的过程发现每个状态都跟0状态有关系,但是dp[0]不确定,但是幸运的是这是一个线性变换,所以状态转移的时候记录一下dp[0]的系数,最后移项输出就好了. dp[i]=dp[i+x]*(k1*k2*k3);(x=i+j…

题目:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4530 dp[i]表示现在存在i个吸血鬼要达成目标(全为吸血鬼)天数的数学期望假如现在再增加一天,这一天可能会增加一个吸血鬼,p1*(dp[i+1]+1)表示接下来的一天增加了一个吸血鬼,所以为(dp[i+1]+1),还有一种可能就是没有增加吸血鬼,概率自然是(1-p1)dp[i]+1表示接下来的一天没有增加吸血鬼,但向后推移了一天因此dp[i]这个状态可以转移到dp[i…