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搜索关键字:vitual judge    ( 1412个结果
hlg1306再遇攻击--射线法判断点是否在多边形内部
再遇攻击Time Limit: 1000 MSMemory Limit: 65536 KTotal Submit: 313(40 users)Total Accepted: 91(32 users)Rating:Special Judge:NoDescriptionDota中英雄技能攻击会有一个范围...
分类:其他好文   时间:2014-07-22 23:08:13    阅读次数:423
hdu 1019 最小公倍数
简单题 注意__int64 的使用Problem : 1019 ( Least Common Multiple ) Judge Status : AcceptedRunId : 10599776 Language : C++ Author : xiaoniuwinCode Ren...
分类:其他好文   时间:2014-07-22 23:06:34    阅读次数:311
PAT 1075. PAT Judge (25)
PAT 1075. PAT Judge (25)
分类:其他好文   时间:2014-05-07 18:10:40    阅读次数:242
193 - Graph Coloring(DFS)
题目:193 - Graph Coloring 题目大意:给出一个图,图里面有点和边,要求相邻的点不可以都是黑色的,问怎样上色黑色的点最多的,给出字典序最大的那种组合情况。#include #include const int N = 105; int n, m, s[N][N], ans[N], cas, count, vis[N]; bool judge (int x, in...
分类:其他好文   时间:2014-05-03 17:22:31    阅读次数:282
poj1598
Excuses, Excuses! Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3615   Accepted: 1254 Description Judge Ito is having a problem with people subpoenaed fo...
分类:其他好文   时间:2014-05-03 17:10:03    阅读次数:278
POJ 2676 数码问题DLX
Sudoku Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13023   Accepted: 6455   Special Judge Description Sudoku is a very simple task. A square table with 9...
分类:其他好文   时间:2014-05-03 00:31:30    阅读次数:400
UVA之409 - Excuses, Excuses!
Excuses, Excuses!  Judge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time requi...
分类:其他好文   时间:2014-05-02 23:54:43    阅读次数:474
HDU 2828 DLX搜索
Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 771    Accepted Submission(s): 230 Special Judge Problem Description There are seve...
分类:其他好文   时间:2014-05-02 02:38:38    阅读次数:443
求一个数的最大素因子(python实现)
首先想到的是,将这个数进行素因子分解,得到所有的因子,然后取最大的。 首先写一个判断一个数是否是素数的方法: #judge a number whether a prime def judgePrime(self,number,pme): if number < 2: ...
分类:编程语言   时间:2014-05-01 09:05:26    阅读次数:3333
hdu1221
求矩形和圆是否相交或相切。 #include #include #define judge(x,y) x<y||fabs(x-y)<0.00000001 double dis(int x,int y,int a,int b) { return (x-a)*(x-a)+(y-b)*(y-b); } int main() { int T; scanf(...
分类:其他好文   时间:2014-04-30 22:45:38    阅读次数:365
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