A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of characters in a fictional setting. Players take responsibility for acting out these roles within a narrative, either through literal acting or thr…

J - Josephina and RPG Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4800 Description A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of cha…

Josephina and RPG Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 929    Accepted Submission(s): 265Special Judge Problem Description A role-playing game (RPG and sometimes roleplaying game) is…

Josephina and RPG Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 511    Accepted Submission(s): 139Special Judge Problem Description A role-playing game (RPG and sometimes roleplaying game) is…

第一年参加现场赛,比赛的时候就A了这一道,基本全场都A的签到题竟然A不出来,结果题目重现的时候1A,好受打击 ORZ..... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4800 题目大意:给定C(3,N)支队伍之间对战的获胜概率,再给定一个序列存放队伍编号,每次获胜之后可以选择和当前战胜的对手换队伍.问按给定序列依次挑战全部胜利的最大概率. 解题思路:状压DP dp[i][j]表示使用队伍i从编号j开始挑战全胜的概率,ai[i]表示i位置的队…

简单dp #include<cstdio> #define maxn 10005 #include<cstring> #include<algorithm> using namespace std; ][maxn]; ][]; int num[maxn]; int main() { int n,m; while(scanf("%d",&m)!=EOF) { m=(m*(m-)*(m-)/); ; i<m; i++) { ; j<m…

题目传送门 题意:告诉你C(m,3)个队伍相互之间的胜率,然后要你依次对战n个AI队伍,首先任选一种队伍,然后战胜一个AI后可以选择替换成AI的队伍,也可以不换,问你最后最大的胜率是多少. 分析:dp[i][j][0/1] 表示第i个AI,用j的id去攻打,此j可以是上一个状态交换AI的id而来也可以不是,状态转移方程: dp[i][j][0] = max (dp[i-1][j][0], dp[i-1][j][1]) * p[j][a[i]]; if (i > 1) dp[i][a[i-1]][…

Josephina and RPG Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of characters in a fictional setting. Players take responsibil…

RPG的错排 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2068 题目大意: 有N个人对应N个名字,然后你去把每一个名字对应到每个人,只要求答对一半及以上就算是过关,问有多少组答案能让他过关. 思考过程: 一眼就发现是错排,从N个人当中选出M个,然后进行错排,M可以从 0 一直到 N / 2. 我还是wa了,因为一开始的时候没有考虑到M可以为0.如果M为0,即表示没有全部排错,为一组答案. 代码: #include<iostream> #in…

不容易系列之(3)—— LELE的RPG难题 Problem Description 人称“AC女之杀手”的超级偶像LELE最近忽然玩起了深沉,这可急坏了众多“Cole”(LELE的粉丝,即"可乐"),经过多方打探,某资深Cole终于知道了原因,原来,LELE最近研究起了著名的RPG难题: 有排成一行的n个方格,用红(Red).粉(Pink).绿(Green)三色涂每个格子,每格涂一色,要求任何相邻的方格不能同色,且首尾两格也不同色．求全部的满足要求的涂法. 以上就是著名的RPG难题.…