ACM----数学之美
用来存放做ACM过程中遇到的各种数论、计算几何、组合数学,简单数学等等所有与数学有关的问题
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人一我十,人十我万,永不放弃——kuangbin
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GCD Matrix----数论
GCD Matrixlockedby ma5termindProblemSubmissionsLeaderboardDiscussionsEditorial Alex has two arrays defined as and . He created an matrix,原创 2017-02-09 14:08:16 · 670 阅读 · 0 评论 -
Widget Factory--高斯消元
Widget FactoryTime Limit: 7000MS Memory Limit: 65536KTotal Submissions: 5806 Accepted: 2013DescriptionThe widget factory produces several different kinds of wid原创 2016-12-07 01:06:46 · 476 阅读 · 0 评论 -
SETI--高斯消元
SETITime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 2030 Accepted: 1236DescriptionFor some years, quite a lot of work has been put into listening to e原创 2016-12-11 21:46:40 · 356 阅读 · 0 评论 -
Flip Game--高斯消元
Flip GameTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42090 Accepted: 18169DescriptionFlip game is played on a rectangular 4x4 field with two-sided原创 2016-12-26 13:21:11 · 420 阅读 · 0 评论 -
EXTENDED LIGHTS OUT--高斯消元
EXTENDED LIGHTS OUTTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9807 Accepted: 6360DescriptionIn an extended version of the game Lights Out, is a p原创 2016-12-26 13:29:03 · 358 阅读 · 0 评论 -
You can Solve a Geometry Problem too----判断两线段是否相交
快速排斥实验+跨立实验先来讲解一下题目需要用到的前置技能1)快速排斥实验2)跨立实验先来讲快速排斥吧快速排斥可以理解为两条线段相交的“预处理”,先大体的判断一下,如果最开始最简单的判断都过不了那么两条线段一定不相交,快速排斥实验,就是以两条线段为对角线,建立两个矩形,如果两个矩形不相交,那么两条线段一定不相交,如果两个矩形相交了,还需要经过跨立实验来判断两条线段是不是相原创 2016-05-23 16:49:46 · 826 阅读 · 2 评论 -
Beauty Contest--凸包
Beauty ContestTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 34687 Accepted: 10728DescriptionBessie, Farmer John's prize cow, has just won first plac原创 2016-08-05 15:49:50 · 502 阅读 · 0 评论 -
Wall--凸包
WallTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 35059 Accepted: 11945DescriptionOnce upon a time there was a greedy King who ordered his chief Arc原创 2016-08-05 16:00:18 · 373 阅读 · 0 评论 -
Expanding Rods--二分答案
Expanding RodsTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 15096 Accepted: 4009DescriptionWhen a thin rod of length L is heated n degrees, it expan原创 2016-08-10 21:44:37 · 460 阅读 · 0 评论 -
Pie--二分答案
PieTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15241 Accepted: 5217 Special JudgeDescriptionMy birthday is coming up and traditionally I'm ser原创 2016-08-11 21:00:53 · 469 阅读 · 0 评论 -
Pipe--计算几何叉积的应用
PipeTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10307 Accepted: 3191DescriptionThe GX Light Pipeline Company started to prepare bent pipes for the原创 2016-08-12 14:42:25 · 577 阅读 · 0 评论 -
Fishnet--计算几何
FishnetTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2132 Accepted: 1343DescriptionA fisherman named Etadokah awoke in a very small island. He could原创 2016-08-19 11:45:48 · 546 阅读 · 0 评论 -
Disks--计算几何
DisksTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 1652 Accepted: 395DescriptionConsider N floating point numbers N representing the radii of N disk原创 2016-09-14 15:59:55 · 559 阅读 · 0 评论 -
count_prime----容斥+唯一分解定理
题目链接:https://icpc.njust.edu.cn/Contest/749/C/此题是容斥+唯一分解定理,建议先去学习前置技能,这两个知识点都很好理解,适当的去学一下唯一分解定理的扩展应用(虽说到现在我还没有做到关于他应用的题=-=)。求一个区间中有多少个数和给定的数字互斥(互质),需要对给定的数字进行素因数分解,然后利用二进制0 1的特性去组合每一个数字,最后利用容斥的奇加原创 2016-05-09 19:00:25 · 1003 阅读 · 1 评论 -
青蛙的约会----扩展欧几里得+乘法逆元
青蛙的约会Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 103947 Accepted: 20246Description两只青蛙在网上相识了,它们聊得很开心,于是觉得很有必要见一面。它们很高兴地发现它们住在同一条纬度线上,于是它们约定各自朝西跳,直到碰原创 2016-05-13 19:03:18 · 1162 阅读 · 0 评论 -
分解素因子----唯一分解定理
分解素因子Time Limit: 1500ms Memory limit: 10000K 有疑问?点这里^_^题目描述假设x是一个正整数,它的值不超过65535(即1输入输入的第一行含一个正整数k (1输出每个测试例对应一行输出,输出x的素数乘积表示式,式中的素数从小到大排列,两个素数之间用“*”表示乘法原创 2016-05-25 20:46:51 · 8615 阅读 · 0 评论 -
GCD问题--莫比乌斯反演
GCD问题Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^题目描述给出区间 gcd(x,y) = k,a 。T(1 T行,每行五个整数输出 输出示例输入2384 31270 278 37299 70336 35722 493 16985原创 2016-08-03 22:00:38 · 1482 阅读 · 0 评论 -
The Embarrassed Cryptographer--高精度
The Embarrassed CryptographerTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13999 Accepted: 3819DescriptionThe young and very promising cryptographer原创 2016-08-06 21:52:22 · 1090 阅读 · 3 评论 -
Sumdiv--数论+快速幂取模+唯一分解定理+欧拉筛
SumdivTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 18987 Accepted: 4767DescriptionConsider two natural numbers A and B. Let S be the sum of all nat原创 2016-08-08 11:51:09 · 776 阅读 · 0 评论 -
C Looooops--模线性方程
C LooooopsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23388 Accepted: 6443DescriptionA Compiler Mystery: We are given a C-language style for loop原创 2016-08-08 21:25:58 · 541 阅读 · 0 评论 -
Number Sequence--组合数学
Number SequenceTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 37890 Accepted: 10952DescriptionA single positive integer i is given. Write a program t原创 2016-08-13 21:49:02 · 315 阅读 · 0 评论 -
Code--组合数学
CodeTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9320 Accepted: 4454DescriptionTransmitting and memorizing information is a task that requires diff原创 2016-08-15 16:33:59 · 396 阅读 · 0 评论 -
Round Numbers--杨辉三角,组合数学
转载地址:http://www.cnblogs.com/lyy289065406/archive/2011/07/31/2122758.html大致题意:输入两个十进制正整数a和b,求闭区间 [a ,b] 内有多少个Round number所谓的Round Number就是把一个十进制数转换为一个无符号二进制数,若该二进制数中0的个数大于等于1的个数,则它就是一个Round Numbe转载 2016-08-18 21:35:04 · 871 阅读 · 0 评论 -
Strange Way to Express Integers--扩展欧几里得和中国剩余定理
Strange Way to Express IntegersTime Limit: 1000MS Memory Limit: 131072KTotal Submissions: 14124 Accepted: 4570DescriptionElina is reading a book written by Ru原创 2016-08-20 02:15:41 · 392 阅读 · 0 评论 -
E. Timofey and remoduling----数论及数学公式
E. Timofey and remodulingtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLittle Timofey likes integers a l原创 2017-02-07 17:11:48 · 1046 阅读 · 0 评论
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