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SPOJ 7001. Visible Lattice Points （莫比乌斯反演）

摘要： 7001. Visible Lattice PointsProblem code: VLATTICEConsider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible from point Y iff no other lattice point lies on the segment joining X and Y.Input 阅读全文

posted @ 2013-08-21 21:14 kuangbin 阅读(1655) 评论(2) 推荐(0) 编辑

BZOJ 2301: [HAOI2011]Problem b （莫比乌斯反演）

摘要： 2301: [HAOI2011]Problem bTime Limit:50 SecMemory Limit:256 MBSubmit:436Solved:187[Submit][Status]Description对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。Input第一行一个整数n，接下来n行每行五个整数，分别表示a、b、c、d、kOutput共n行，每行一个整数表示满足要求的数对(x,y)的个数Sample Input22 5 1 5 11 5 1 5 2 Sample Output1 阅读全文

posted @ 2013-08-21 21:12 kuangbin 阅读(4636) 评论(0) 推荐(0) 编辑

HDU 1695 GCD （莫比乌斯反演）

摘要： GCDTime Limit: 6000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4291Accepted Submission(s): 1502Problem DescriptionGiven 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x an 阅读全文

posted @ 2013-08-21 21:10 kuangbin 阅读(8238) 评论(0) 推荐(2) 编辑

HDU 4691 Front compression （2013多校9 1006题后缀数组）

摘要： Front compressionTime Limit: 5000/5000 MS (Java/Others)Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 158Accepted Submission(s): 63Problem DescriptionFront compression is a type of delta encoding compression algorithm whereby common prefixes and their lengths are recorded so that th 阅读全文

posted @ 2013-08-20 19:32 kuangbin 阅读(1780) 评论(0) 推荐(0) 编辑

HDU 4690 EBCDIC （2013多校 1005题胡搞题）

摘要： EBCDICTime Limit: 2000/2000 MS (Java/Others)Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 160Accepted Submission(s): 81Problem DescriptionA mad scientist found an ancient message from an obsolete IBN System/360 mainframe. He believes that this message contains some very important s 阅读全文

posted @ 2013-08-20 19:31 kuangbin 阅读(938) 评论(0) 推荐(0) 编辑

HDU 4686 Arc of Dream （2013多校9 1001 题，矩阵）

摘要： Arc of DreamTime Limit: 2000/2000 MS (Java/Others)Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 0Accepted Submission(s): 0Problem DescriptionAn Arc of Dream is a curve defined by following function:wherea0 = A0ai = ai-1*AX+AYb0 = B0bi = bi-1*BX+BYWhat is the value of AoD(N) modulo 1, 阅读全文

posted @ 2013-08-20 19:28 kuangbin 阅读(1434) 评论(1) 推荐(1) 编辑

HDU 1695 GCD （欧拉函数+容斥原理)

摘要： GCDTime Limit: 6000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4272Accepted Submission(s): 1492Problem DescriptionGiven 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x an 阅读全文

posted @ 2013-08-19 23:25 kuangbin 阅读(2825) 评论(0) 推荐(0) 编辑

【转】[专题学习][计算几何]

摘要：原文地址：http://www.cnblogs.com/ch3656468/archive/2011/03/02/1969303.html基本的叉积、点积和凸包等东西就不多说什么了，网上一搜一大堆，切一些题目基本熟悉了就差不多了。一些基本的题目可以自己搜索，比如这个blog：http://blog.sina.com.cn/s/blog_49c5866c0100f3om.html 接下来，研究了半平面交，思想方法看07年朱泽园的国家队论文，模板代码参考自我校大牛韬哥：http://www.owent.net/2010/10/acm-%E8%AE%A1%E7%AE%97%E5%87%A0%E.. 阅读全文

posted @ 2013-08-19 12:42 kuangbin 阅读(3059) 评论(0) 推荐(3) 编辑

POJ 1755 Triathlon （半平面交）

摘要： TriathlonTime Limit:1000MSMemory Limit:10000KTotal Submissions:4733Accepted:1166DescriptionTriathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the th 阅读全文

posted @ 2013-08-19 00:37 kuangbin 阅读(1532) 评论(0) 推荐(0) 编辑

POJ 3384 Feng Shui （半平面交）

摘要： Feng ShuiTime Limit:2000MSMemory Limit:65536KTotal Submissions:3743Accepted:1150Special JudgeDescriptionFeng shui is the ancient Chinese practice of placement and arrangement of space to achieve harmony with the environment. George has recently got interested in it, and now wants to apply it to his 阅读全文

posted @ 2013-08-18 16:19 kuangbin 阅读(1166) 评论(0) 推荐(0) 编辑