题目链接 LOOPS Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 2630    Accepted Submission(s): 1081 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to h…

LOOPS Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 8453    Accepted Submission(s): 3397 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h…

简单的概率DP入门题 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #define MAX 1003 using namesp…

题目链接 495. Kids and Prizes Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the company organiz…

首先要推出dp[i]的期望方程,会发现每一项都和dp[0]相关, 那我们将dp[i]设为和dp[0]有关的式子dp[i]=a[i]*dp[0]+b[i],然后再回代到原来的期望方程里 然后进行整理,可以发现两个系数a[i],b[i]是可以逆推的,并且通过求出a[0],b[0]可以求出dp[0] #include<bits/stdc++.h> using namespace std; #define maxn 1050 double A[maxn],B[maxn],p[maxn]; int ma…

题目链接 Card Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2711    Accepted Submission(s): 1277Special Judge Problem Description In your childhood, do you crazy for collecting the beaut…

D - LOOPS Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help her friend Madoka save the world. But because of the plot…

题意:就是让你从(1,1)走到(r, c)而且每走一格要花2的能量,有三种走法:1,停住.2,向下走一格.3,向右走一格.问在一个网格中所花的期望值. 首先:先把推导动态规划的基本步骤给出来. · 1.设变量:(注意:设置变量时,要能够使整个求解过程可以分为多个阶段.) 2.分析阶段决策,并写出决策函数.(也就是能体现前阶段决策后阶段关系的函数) 3.写出指标函数.(也是就是我们得出解的函数.) 先第一步:设置变量,我们分析这个题的是从(1,1)到(r, c)那么什么能体现“阶段”这个词的东西呢…

在拐~ #include <stdio.h> #include <cstring> #include <iostream> #include <map> #include <cmath> template <class T> inline bool rd(T &ret) { char c; int sgn; if(c=getchar(),c==EOF) return 0; while(c!='-'&&(c<…

HDU 3853    LOOPS 题目大意是说人现在在1,1,需要走到N,N,每次有p1的可能在元位置不变,p2的可能走到右边一格,有p3的可能走到下面一格,问从起点走到终点的期望值 这是弱菜做的第一道概率DP的题,首先是看了一下有关概率DP的资料,大概知道一般球概率就是从起点推到终点,求期望就是从终点推到起点 考虑这题的做法,其实很简单设DP[i][j]表示从i,j到达终点所需时间的期望值 DP[i][j] =p1 *  DP[i][j] + p2 * DP[i][j+1] + p3 * D…