five专题

XTOJ 1173 Five Tiger

OJ题目:click here ~~ AC_CODE int grid[6][6];int dawuhu(){return grid[1][1]&& grid[1][5]&& grid[5][1]&& grid[5][5]&& grid[3][3];}int wuhu(){int ret = 0;for(int i = 1;i <= 3;i++)for(int j = 1;j <=

BestCoder Round #62 (div.2)Clarke and five-pointed star(极角排序,判断五边形)

题目链接 题意:给你五个点,问这五个点是否可以组成正五边形(正五角星,等价于正五边形)。 解答:先极角排序,(让五个点按照顺时针或者逆时针的顺序)然后我们计算五条边是不是一样,然后在看对角线是不是都一样。 #include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<set>#includ

docker 数据存储(thirty-five day)

1、打包 [root@docker1 ~]# docker save -o centos.tar centos:latest[root@docker1 ~]# systemctl start docker[root@docker1 ~]# docker ps -allCONTAINER ID IMAGE COMMAND CREATED

HikariCP closes connections in five cases

HikariCP determined that the connection was dead, i.e. (connection is evicted or dead), and therefore attempted to close it. The driver then said, “Sorry the connection is already closed”, which is no

【Canvas与艺术】绘制铜质蓝底五周年(Five Years Anniversary)徽章

【关键点】 利用二次贝塞尔曲线生成环形波纹轮廓。 【成果图】 【代码】 <!DOCTYPE html><html lang="utf-8"><meta http-equiv="Content-Type" content="text/html; charset=utf-8"/><head><title>铜质五周年徽章</title><style type="text/css

蓝桥杯备战刷题five(自用)

1.数字三角形(方向次数限制,动态规划)   //如果n为奇数时,最后必然走到最后行最中间的数,如果为偶数,则取中间两个数的最大值,   //因为向左下走的次数与向右下走的次数相差不能超过 1 #include <iostream>using namespace std;const int N=110;int g[N][N];int f[N][N];int n;int ans;

CodeForce 825B Five-In-a-Row简单模拟

B. Five-In-a-Row time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alice and Bob play 5-in-a-row game. They have a playi

Five-unity3D火箭鼠小游戏

今天的游戏可以玩了,仍存在的问题: 1.激光不关闭 2.飞鼠下落状态并不是fall 3.有“!”错误提示 新内容: 1.飞鼠的死亡模式:碰到障碍物会死亡(图片-动画) 2.奖励模式:金币的随机放置(图层、脚本编写) 3.记录模式:屏幕角上记录获取到的金币数量(GUI、文字美化) 感想:今天是补课的倒数第二天,一个跑酷类小游戏已经大体成型,想了想如果把时间挤在一起做的话大概需要个一两

HDU 5562 Clarke and five-pointed star (BC)

Clarke and five-pointed star Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other) Total Submission(s) : 8   Accepted Submission(s) : 2 Font: Times New Roman | Verd

Five in a Row, Again-记录状态的dfs+剪枝

用一个二进制记录当前状态。 二进制中的0代表当前点已经被走过,1代表当前点还未走过。 如果当前二进制之前计算过,且值大于当前值的话,那么就终止不走。 #include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int shuyu[5001];int n;

concurrent five things that you didn't know

http://www.ibm.com/developerworks/cn/java/j-5things4.html