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…

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…

Problem Description Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an M×N matrix. The wall has M×N squares in all. In the whole problem we denotes (x,y) to be the square at the x-th row, y-th…

题目链接: J - Joyful  HDU - 5245 题目大意:给你一个n*m的矩阵,然后你有k次涂色机会,然后每一次可以选定当前矩阵的一个子矩阵染色,问你这k次用完之后颜色个数的期望. 具体思路:颜色个数的期望等于每一个方块单独的期望加起来,就是总的期望. 对于当前的方块的期望,我们先计算这个方块不会出现的概率,就是当前的(x,y),先计算出当前的两个点在他周围四整块的出现的概率,但是这样四个角会重复计算,再去掉就好了. AC代码: #include<bits/stdc++.h> usi…

http://acm.hdu.edu.cn/showproblem.php?pid=5245 题意: 给出一个n*m的矩阵格子,现在有k次操作,每次操作随机选择两个格子作为矩形的对角,然后将这范围内的格子填色,现在要求经过k次操作后填色格子的期望值. 思路: 给个格子都是独立的,所以只需要计算出每个格子经过k次操作后被填色的概率即可,最后所有格子相加就是期望值.但是直接求填色概率不好求,求不被填色概率会比较容易.假设一次操作的不被填色概率为p,那么k次之后的概率为p^k,最后该格子填色概率就是1…

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…

题意:进行K次染色,每次染色会随机选取一个以(x1,y1),(x2,y2)为一组对角的子矩阵进行染色,求K次染色后染色面积的期望值(四舍五入). 析:我们可以先求出每个格子的期望,然后再加起来即可.我们可以把格子进行划分,然后再求概率. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cst…

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…