题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5159 题解: 考虑没一个数的贡献,一个数一次都不出现的次数是(x-1)^b,而总的排列次数是x^b,所以每一个数有贡献的次数都是x^b-(x-1)^b,所以最后推导的公式就是: (x^b-(x-1)^b)*(1+2+...+x)/(x^b)=(1-((x-1)/x)^b)*(1+x)*x/2 代码: #include<iostream> #include<cstdio> #inclu…

Problem Description There are x cards on the desk, they are numbered from 1 to x. The score of the card which is numbered i(1<=i<=x) is i. Every round BieBie picks one card out of the x cards,then puts it back. He does the same operation for b round…

B - Card Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description There are x cards on the desk, they are numbered from 1 to x. The score of the card which is numbered i(1<=i<=x) is i. Every round Bi…

Card Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 191    Accepted Submission(s): 52Special Judge Problem Description There are x cards on the desk, they are numbered from 1 to x. The score o…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4336 Card Collector Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) 问题描述 In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that,…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5159 题解:假设在 x 张牌中选b张牌,那么有 x^b 种选法,如果在 (x-1) 张牌中选 b 张牌,那么有 (x-1)^b 种选法,所以第 i 张牌出现的概率是 (x^b-(x-1)^b)/x^b 再对每张牌乘上牌面的值即是期望. #include<stdio.h> #include<iostream> #include<string.h> #include <…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4336 题目大意:有n种卡片,需要吃零食收集,打开零食,出现第i种卡片的概率是p[i],也有可能不出现卡片.问你收集齐n种卡片,吃的期望零食数是多少? 状态压缩:f[mask],代表收集齐了mask,还需要吃的期望零食数. 打开包装,有3种情况,第一种:没有卡片,概率(1-sigma(p[i])) 第二种,在已知种类中:概率sigma(p[j]) 第三种,在未知种类中:p[k] 因此 f[mask]…

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=4336 题意简单,直接用容斥原理即可 AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cstdlib> #include <cmath> #include <vector> #include &…

Problem Description In your childhood, people in the famous novel Water Margin, you will win an amazing award. As a smart boy, you notice that to win the award, you must buy much more snacks than it seems to be. To convince your friends not to waste…

题目:点击打开链接 题意:两个人纸牌游戏,牌大的人得分.牌大:2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < T < J < Q < K < A .值一样看花色, hearts (红心) > spades (黑桃) > diamond (方块) > clubs (梅花).问Eve 能得多少分.(每次得1分) 分析:将Eve的每张牌与Adam的所有牌比较,与所有比这张牌小的Adam的牌连边. 然后求最大…