05 2016 档案

摘要:map映照容器的元素数据是一个键值和一个映照数据组成的,键值与映照数据之间具有一一映照的关系。 map映照容器的数据结构是采用红黑树来实现的,插入键值的元素不允许重复,比较函数只对元素的键值进行比较,元素的各项数据可通过键值检索出来。 使用map容器需要头文件包含语句“#include<map>”。 阅读全文
posted @ 2016-05-30 20:34 Kiven#5197 阅读(424) 评论(3) 推荐(0) 编辑
摘要:近期学习了STL中set的使用,在此写一点点总结和自己的一些体悟。 set集合容器实现了红黑树(Red-Black Tree)的平衡二叉检索树的的数据结构,在插入元素时,它会自动调整二叉树的排列,把该元素放到适当的位置,以确保每个子树根节点的键值大于左子树所有节点的键值,而小于右子树所有节点的键值; 阅读全文
posted @ 2016-05-27 22:55 Kiven#5197 阅读(275) 评论(0) 推荐(0) 编辑
摘要:水果 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5303 Accepted Submission(s): 2022 Problem Desc 阅读全文
posted @ 2016-05-26 21:33 Kiven#5197 阅读(227) 评论(0) 推荐(0) 编辑
摘要:Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 103000 Accepted Submission(s 阅读全文
posted @ 2016-05-26 19:58 Kiven#5197 阅读(281) 评论(0) 推荐(0) 编辑
摘要:统计难题 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)Total Submission(s): 30259 Accepted Submission(s): 11770 Problem 阅读全文
posted @ 2016-05-26 19:21 Kiven#5197 阅读(245) 评论(0) 推荐(0) 编辑
摘要:C. Pearls in a Row time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output C. Pearls in a Row ti 阅读全文
posted @ 2016-05-26 19:13 Kiven#5197 阅读(427) 评论(0) 推荐(0) 编辑
摘要:{A} + {B} Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17014 Accepted Submission(s): 7110 Pro 阅读全文
posted @ 2016-05-24 22:36 Kiven#5197 阅读(152) 评论(0) 推荐(0) 编辑
摘要:10954 - Add All Time limit: 3.000 seconds Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescend 阅读全文
posted @ 2016-05-24 16:03 Kiven#5197 阅读(427) 评论(0) 推荐(0) 编辑
摘要:Card Stacking Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3927 Accepted: 1541 Description Bessie is playing a card game with her N-1 (2 阅读全文
posted @ 2016-05-24 15:26 Kiven#5197 阅读(488) 评论(0) 推荐(0) 编辑
摘要:神、上帝以及老天爷 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32906 Accepted Submission(s): 13449 Pro 阅读全文
posted @ 2016-05-22 19:06 Kiven#5197 阅读(344) 评论(0) 推荐(0) 编辑
摘要:放苹果 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 30213 Accepted: 19053 Description 把M个同样的苹果放在N个同样的盘子里,允许有的盘子空着不放,问共有多少种不同的分法?(用K表示)5,1,1 阅读全文
posted @ 2016-05-22 16:55 Kiven#5197 阅读(241) 评论(0) 推荐(0) 编辑
摘要:E - Jolly Jumpers Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status E - Jolly Jumpers Submit Status Description A sequence 阅读全文
posted @ 2016-05-20 23:42 Kiven#5197 阅读(299) 评论(0) 推荐(0) 编辑
摘要:阿牛的EOF牛肉串 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29999 Accepted Submission(s): 14095 Pro 阅读全文
posted @ 2016-05-20 22:27 Kiven#5197 阅读(300) 评论(0) 推荐(0) 编辑
摘要:今年暑假不AC Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44625 Accepted Submission(s): 23870 Probl 阅读全文
posted @ 2016-05-20 17:21 Kiven#5197 阅读(268) 评论(0) 推荐(0) 编辑
摘要:冒泡排序(Bubble Sort) 冒泡排序(Bubble Sort),是一种计算机科学领域的较简单的排序算法。 它重复地走访过要排序的数列,一次比较两个元素,如果他们的顺序错误就把他们交换过来。走访数列的工作是重复地进行直到没有再需要交换,也就是说该数列已经排序完成。 这个算法的名字由来是因为越大 阅读全文
posted @ 2016-05-14 21:32 Kiven#5197 阅读(359) 评论(0) 推荐(0) 编辑
摘要:C - Present Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 460C C - Present Submit Status Pra 阅读全文
posted @ 2016-05-14 20:39 Kiven#5197 阅读(323) 评论(0) 推荐(0) 编辑
摘要:B - Little Dima and Equation Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 460B B - Little D 阅读全文
posted @ 2016-05-14 16:44 Kiven#5197 阅读(187) 评论(0) 推荐(0) 编辑
摘要:A - Vasya and Socks Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 460A A - Vasya and Socks S 阅读全文
posted @ 2016-05-14 16:39 Kiven#5197 阅读(261) 评论(0) 推荐(0) 编辑
摘要:Number Steps Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4987 Accepted Submission(s): 3030 Pr 阅读全文
posted @ 2016-05-12 21:57 Kiven#5197 阅读(136) 评论(0) 推荐(0) 编辑
摘要:Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 67912 Accepted Submission(s): 21226 阅读全文
posted @ 2016-05-12 20:57 Kiven#5197 阅读(196) 评论(0) 推荐(0) 编辑
摘要:Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 103000 Accepted Submission(s 阅读全文
posted @ 2016-05-12 20:13 Kiven#5197 阅读(280) 评论(0) 推荐(0) 编辑
摘要:Fibonacci Again Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 50987 Accepted Submission(s): 241 阅读全文
posted @ 2016-05-11 23:09 Kiven#5197 阅读(222) 评论(0) 推荐(0) 编辑
摘要:sort Time Limit: 6000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40503 Accepted Submission(s): 11823 Problem 阅读全文
posted @ 2016-05-11 21:58 Kiven#5197 阅读(193) 评论(0) 推荐(0) 编辑
摘要:前m大的数 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15717 Accepted Submission(s): 5362 Problem 阅读全文
posted @ 2016-05-11 20:55 Kiven#5197 阅读(254) 评论(0) 推荐(0) 编辑
摘要:排序 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 48878 Accepted Submission(s): 14110 Problem De 阅读全文
posted @ 2016-05-11 20:21 Kiven#5197 阅读(366) 评论(0) 推荐(0) 编辑
摘要:今晚翻出了之前收藏的一个有趣的问题--一千个苹果的分装: 现在有1000个苹果,和10个箱子,如何把这1000个苹果装在这10个箱子里,才能使不管任何数量(1-1000)的苹果,都能一次给出? 典型的二进制数变种,1000个苹果,最接近1024,转化为2进制,需要占用10个bit,则从右往左数,第一 阅读全文
posted @ 2016-05-09 23:01 Kiven#5197 阅读(1222) 评论(0) 推荐(0) 编辑
摘要:母猪的故事 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9222 Accepted Submission(s): 5699 Problem D 阅读全文
posted @ 2016-05-09 21:56 Kiven#5197 阅读(214) 评论(0) 推荐(0) 编辑
摘要:The sum problem Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21744 Accepted Submission(s): 639 阅读全文
posted @ 2016-05-09 21:26 Kiven#5197 阅读(358) 评论(0) 推荐(0) 编辑
摘要:不容易系列之(3)—— LELE的RPG难题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 43557 Accepted Submission( 阅读全文
posted @ 2016-05-09 20:27 Kiven#5197 阅读(266) 评论(0) 推荐(0) 编辑
摘要:人见人爱A-B Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 66259 Accepted Submission(s): 18470 Probl 阅读全文
posted @ 2016-05-09 19:51 Kiven#5197 阅读(207) 评论(0) 推荐(0) 编辑
摘要:A == B ? Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 88211 Accepted Submission(s): 13922 Prob 阅读全文
posted @ 2016-05-09 15:32 Kiven#5197 阅读(361) 评论(0) 推荐(0) 编辑
摘要:改革春风吹满地 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28794 Accepted Submission(s): 14761 Probl 阅读全文
posted @ 2016-05-09 14:48 Kiven#5197 阅读(198) 评论(0) 推荐(0) 编辑
摘要:F - 棋盘问题 Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1321 F - 棋盘问题 Submit Status Practice POJ 1321 阅读全文
posted @ 2016-05-06 23:46 Kiven#5197 阅读(300) 评论(0) 推荐(0) 编辑
摘要:B - Simple Game Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 570B B - Simple Game Submit St 阅读全文
posted @ 2016-05-06 23:39 Kiven#5197 阅读(602) 评论(0) 推荐(0) 编辑
摘要:A - Elections Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u SubmitStatusPracticeCodeForces 570A A - Elections SubmitStatusPrac 阅读全文
posted @ 2016-05-06 23:13 Kiven#5197 阅读(328) 评论(0) 推荐(0) 编辑
摘要:{A} + {B} Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17014 Accepted Submission(s): 7110 Pro 阅读全文
posted @ 2016-05-05 21:29 Kiven#5197 阅读(199) 评论(0) 推荐(0) 编辑
摘要:仙人球的残影 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6851 Accepted Submission(s): 3192 Problem 阅读全文
posted @ 2016-05-04 23:57 Kiven#5197 阅读(165) 评论(0) 推荐(0) 编辑
摘要:Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line 阅读全文
posted @ 2016-05-04 23:35 Kiven#5197 阅读(444) 评论(1) 推荐(0) 编辑