Description Anton and Dasha like to play different games during breaks on checkered paper. By the 11th grade they managed to play all the games of this type and asked Vova the programmer to come up with a new game. Vova suggested to them to play a ga…

com.yuming.YiZhiFanPai01--前拼后广--益智早教游戏记忆翻牌cym…

Spiral Maximum 题目连接: http://codeforces.com/problemset/problem/173/C Description Let's consider a k × k square, divided into unit squares. Please note that k ≥ 3 and is odd. We'll paint squares starting from the upper left square in the following orde…

题目:http://codeforces.com/contest/398/problem/B 有点似曾相识的感觉,记忆中上次那个跟这个相似的 我是用了 暴力搜索过掉的,今天这个肯定不行了,dp方程想了非常久也没有想出来,有点无从下手的感觉,最后还是尝试了一下记忆化搜索,以dp[0][0]为边界,dp[x][y]代表当前有x行y列没有彩色的 瓷砖,搜索起来思路还是非常清晰的,可惜最后那个 算期望公式给写错了,瞎了好久,做出以后看了一下别人的做法,确实有点难想到,把涂好的都放在右下角,这样就仅仅有四…

点我进入题目 题目大意:n个小孩围一圈传球,每个人可以给左边的人或右边的人传球,1号小孩开始,一共传m次,请问有多少种可能的路径使球回到1号小孩. 输入输出:输入n,m,输出路径的数量. 数据范围:40% 3<=n<=30 1<=m<=20 100% 3<=n<=30 1<=m<=30 我是这么想的:膜拟过程,从1号小孩开始dfs,然后加一个记忆化搜索节省时间. dfs(num,tim)表示球传到第num个小孩,已经传过tim次时候,d[num][tim]表…

链接 Codeforces 667C Reberland Linguistics 题意 给你一个字符串,除去前5个字符串后,使剩下的串有长度为2或3的词根组成,相邻的词根不能重复.找到所有的词根 思路 去掉前5个字符,将剩下的串反过来进行记忆化,用vis[last][pos]记录一下当前状态是否做过.last是之前与之相邻的词根.比赛的时候只用了vis[i]错了. 代码 #include <iostream> #include <cstdio> #include <vecto…

Description A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward"). Y…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game ju…

1. 游戏功能和流程图 实现功能:翻开两个一样的牌子就显示,全部翻开游戏结束,设置5种图形,7种颜色,游戏开始提示随机8个牌子 游戏流程图 2. 游戏配置 配置游戏目录 配置游戏(game_conf.py) FPS=30 #游戏帧数WINDOW_WIDTH=640 #窗口大小WINDOW_HEIGHT=480REVEAL_SPEAD=3 #箱子显示的速度BOX_SIZE=40 #箱子的大小宽高相等的正方形GAP_SIZE=10 #箱子之间的间隔BOARD_WIDTH=10 #图标的列数BOARD…

题目链接:https://www.tyvj.cn/Problem_Show.aspx?id=1014 f[i][j]表示区间［i,j］所得到的最小值. 不断地划分区间,把结果保存起来. #include <cstdio> #include <cstdlib> #include <algorithm> using namespace std; ][];], i, j, n, INF=0x7f7f7f7f; void dfs(int l, int r) { ) {f[l][…