http://acm.hdu.edu.cn/showproblem.php?pid=2131 Problem Description Mickey is interested in probability recently. One day , he played a game which is about probability with mini.First mickey gives a letter and a word to mini.Then mini calculate the pr…

Problem Description Mickey is interested in probability recently. One day , he played a game which is about probability with mini.First mickey gives a letter and a word to mini.Then mini calculate the probability that the letter appears in the word.F…

Probability One Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 379    Accepted Submission(s): 293 Problem Description Number guessing is a popular game between elementary-school kids. Teachers…

A simple probability problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 43    Accepted Submission(s): 14 Problem Description Equally-spaced parallel lines lie on an infinite plane. The sep…

Everything Is Generated In Equal Probability \[ Time Limit: 1000 ms\quad Memory Limit: 131072 kB \] 题意 给出一个 \(N\),以相等的概率生成 \(n\) 且 \(n \in [1, N]\),在以相等的概率生成长度为 \(n\) 的数组,最后将生成的数组扔到 \(CALCULATE\) 函数并返回一个数,问这个数的期望. 思路 先解释一下样例是怎么得来的. 令 \(dp[array]\) 表示…

题意: 给定一个N,随机从[1,N]里产生一个n,然后随机产生一个n个数的全排列,求出n的逆序数对的数量,加到cnt里,然后随机地取出这个全排列中的一个非连续子序列(注意这个子序列可以是原序列),再求出这个子序列的逆序数对,加到cnt里,重复这个过程,直到最后取出的为空. 题解: 先不考虑第一步随机从[1,N]里产生一个n,只考虑n给定的情况,求出了f[n],那么最后的结果就是 $  ans[N]=\frac{\sum_{n=1}^N f[n]}{N} $ 赛时和队友利用找规律法和暴力模拟法推出…

Probability 链接:http://acm.hdu.edu.cn/showproblem.php?pid=2131 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8731    Accepted Submission(s): 4228 Problem Description Mickey is interested in pro…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能超过背包的容量1-P. 在这个条件下,让装入背包的物品的总价值,也就是bag[i].[v]的和最大 bag.v是每一件物品的价值,bag.p是每件物品的体积 像上面这样想是行不通的.下面有解释 这道题麻烦的是概率这东西没法用个循环表示出来,根据我以往的经验,指望着把给出的测试数据乘上一百或者一万这种…

A - Robberies Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2955 Appoint description: Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usu…