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HDU1729 Stone Game (SG函数)
HDU1729 Stone Game (SG函数)...
分类:其他好文   时间:2015-04-01 17:39:36    阅读次数:131
[ISSUE] Pick Stone
源引自己的github.iohttp://awarrior.github.io/pick-stone/DescribeTwo players would like to pick up some from one heap of stones. Assume the quantity of ston...
分类:其他好文   时间:2015-03-30 12:53:31    阅读次数:138
HDU 3555 Bomb(数位DP)
Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names)...
分类:其他好文   时间:2015-03-28 06:29:26    阅读次数:133
(hdu step 7.1.3)Lifting the Stone(求凸多边形的重心)
题目:Lifting the StoneTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 168 Accepted Submission(s): 98 Problem DescriptionThere are many secret opening...
分类:其他好文   时间:2015-03-18 12:32:02    阅读次数:194
SPOJ Problem 40:Lift the Stone
求多边形的重心。。啥都不说了,代公式。。#includedouble x0,y0,x1,y1,x2,y2,s,as,gx,gy;int t,n;int main(){ scanf("%d",&t); while(t--){ scanf("%d",&n); sc...
分类:其他好文   时间:2015-03-16 22:31:38    阅读次数:148
log4j.properties配置详解
stone 的 log4j配置详解Log4J的配置文件(Configuration File)就是用来设置记录器的级别、存放器和布局的,它可接key=value格式的设置或xml格式的设置信息。通过配置,可以创建出Log4J的运行环境。1. 配置文件Log4J配置文件的基本格式如下: #配置根Log...
分类:其他好文   时间:2015-03-16 15:56:52    阅读次数:151
(简单) POJ 2253 Frogger,Dijkstra。
Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to ...
分类:其他好文   时间:2015-03-15 00:37:42    阅读次数:154
Sand Maker is an Indispensable Part of Sand Production Line
Sand production line is a special kind of production line for sand and stone, which is defined according to the material type. Its finished products m...
分类:其他好文   时间:2015-03-13 20:33:41    阅读次数:148
Frogger
DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visi...
分类:其他好文   时间:2015-03-07 15:32:03    阅读次数:166
poj 1740 A New Stone Game nim变形
题意: 给n堆石子,两人交替,选择一堆石头后先拿去任意颗,再把剩下的放到其他任意堆,最先拿完所有石子赢,问先手必胜还是必败。 分析; 解决此类问题的一种的思路是先构造策略,然后判断此策略能否满足1.必胜态可到必败态。2.必败态无法到必败态。 代码: //poj 1740 //sep9 #include #include using namespace std; const int m...
分类:其他好文   时间:2015-02-24 09:07:39    阅读次数:142
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