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…

LOOPS Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1864    Accepted Submission(s): 732 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h…

Collecting Bugs Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers…

借鉴自:https://www.cnblogs.com/keyboarder-zsq/p/6216762.html 题意:n个格子,每个格子有一个值.从1开始,每次扔6个面的骰子,扔出几点就往前几步,然后把那个格子的金子拿走: 如果扔出的骰子+所在位置>n,就重新扔,直到在n: 问取走这些值的期望值是多少 解析: [1] [2] [3][4] [5] [6] [7] [8] [9] //格子和值都是一样,所以下述的话,值就是格子,格子就是值... 比如这样的9个格子,我们总底往上来 对于第9个格…

D - Joyful Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an $M \times N$ matrix. The wal…

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…

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…

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…

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>…

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…