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hdu1213
- hdu 1213 边界处理的经典 一定要会做啊 -classic hdu 1213 boundary treatment must do
hdu-3065
- 这是一个关于AC自动机的深入了解的解题报告,能够有效的深入学习AC自动机这个算法-This is a problem solving reports, in-depth understanding of the AC automaton effective in-depth study of the AC automaton algorithm
sanguosha
- hdu 3378 三国杀标程 解题报告-hdu 3378 san guo sha
1097---Lucky-Number
- hdu 1097原来是用链表做的,不想线段树的解法理想一点-hdu 1097
Snooker-Referee
- 解决ZOJ 3426 & HDU 3719 Snooker Referee的问题,实现斯诺克智能裁判的功能。-Solve the problem of ZOJ 3426 & HDU 3719 Snooker Referee snooker intelligent referee function.
HDU
- hdu4425 解题报告,杭电4425结题报告,acm结题报告,杭州电子科技大学结题报告-report hdu4425 problem solving, Hangzhou Xinhua 4425 concluding report, acm concluding report, Hangzhou University of Electronic Science and Technology concluding report
hduacm
- 杭电acm习题(原创ac答案,仅供参考,如有不正,欢迎指教)-hdu acm practice
hduoj
- 杭州电子科技大学acm题库,可以离线观看-hdu practice
HDU1001-Hdu-Girls-Day
- 杭州电子科技大学ACM平台练习题1001解答-The ACM platform Hangzhou University of Electronic Science and Technology practice questions 1001 answers
hdu3666
- hdu 3666的解题报告,是一题差分约束的经典题目-solution to hdu 3666
code
- hdu 1044 Collect More Jewels
HDOJ-icpc
- HDOJ的算法题目,完整的解答方案,含有7个题目的完整程序,希望对你们有帮助-this is the hdu acm,and the algoirth is very perfect
acm
- ACM HDU JAVA语言写的题目 测试通过 可以使用-ACM HDU JAVA language topic to write about the test can be used
hdu-2048
- 神、上帝以及老天爷 输入数据的第一行是一个整数C,表示测试实例的个数,然后是C 行数据,每行包含一个整数n(1<n<=20),表示参加抽奖的人数。 Output 对于每个测试实例,请输出发生这种情况的百分比,每个实例的输出占一行, 结果保留两位小数(四舍五入),-God, God, and God The first line of input data is an integer C, which means that the number of test case
hdu3398
- the problem solved with hdu 3398~-the problem solved with hdu 3398
HDU-2000-2099
- 杭电ACM2000-2099题的解题报告-Hangzhou Xinhua reported problem solving ACM2000-2099' s
HDUP2000-2099
- HDU 2000-2099 解题报告 杭电的2000-2099解题思路-HDU. 2000-2099 the problem solving report hangzhou electric way for 2000-2099
HDU
- 关于矩阵变换,四元数计算的一些实现,结合Phantom设备。-some Utilities about Matrix and Quarternion etc. combing Phantom
HDU-1535-
- 求多源点到单终点的最短路(反向建图),ACM竞赛中应用的小程序。-Seeking multiple sources into a single end-point shortest (reverse construction diagram), ACM contest application applet.
1172
- hdu 1172 猜数字15MS 244K -hdu accepted