Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2060    Accepted Submission(s): 1346 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids lab…

Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1503    Accepted Submission(s): 1025 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids la…

Problem Description Hzz loves aeroplane chess very much. The chess map contains N+ grids labeled to N. Hzz starts at grid . For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are ,,,,,)…

http://acm.hdu.edu.cn/showproblem.php?pid=3345 Problem Description War chess is hh's favorite game:In this game, there is an N * M battle map, and every player has his own Moving Val (MV). In each round, every player can move in four directions as lo…

题意:沿着x轴从0走到大于等于N的某处,每一步的步数由骰子(1,2,3,4,5,6)决定,若恰好走到x轴上某飞行路线的起点,则不计入扔骰子数.问从0走到大于等于N的某处的期望的扔骰子次数. 分析: 1.dp[i]表示从位置i到终点期望的扔骰子次数. 2.很显然倒着往前推,因为从起点0开始,扔骰子的次数有很多种可能,难以计算,但是dp[N]很显然是0,不需要扔骰子即可到达终点. 3.假设当前位于位置i,根据骰子数可能到达的位置有i + j(j=1,2,3,4,5,6),到达其中每个位置的概率都是1…

传送门 期望dp简单题啊. 不过感觉题意不太对. 手过了一遍样例发现如果有捷径必须走. 这样的话就简单了啊. 设f[i]" role="presentation" style="position: relative;">f[i]f[i]表示从第i个格子出发到第n个格子的期望步数. 显然就可以从f[i+1]~f[i+6]转移过来了,注意如果f下标超过n期望步数都是0. 代码: #include<bits/stdc++.h> #define…

//1088(参考博客:http://blog.csdn.net/libin56842/article/details/8950688)//1.编写一个浏览器输入输出(hdu acm1088)://思路:对已经输入的字符串进行处理,遇到<br><hr>分别进行处理.遇到多于80个字符(统计该行的长度)或者<br>或者结束,则换行:遇到<hr>,输出80个'-'.#include<stdio.h>#include<string.h>#d…

http://acm.split.hdu.edu.cn/showproblem.php?pid=4405 Aeroplane chess Problem Description   Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six…

Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1667    Accepted Submission(s): 1123 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids lab…

题意:有一个n个点的飞行棋,问从0点掷骰子(1~6)走到n点须要步数的期望 当中有m个跳跃a,b表示走到a点能够直接跳到b点. dp[ i ]表示从i点走到n点的期望,在正常情况下i点能够到走到i+1,i+2,i+3,i+4,i+5,i+6 点且每一个点的概率都为1/6 所以dp[i]=(dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6])/6  + 1(步数加一). 而对于有跳跃的点直接为dp[a]=dp[b]; #include<stdio.h>…