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RSS订阅原创 指数循环节
指数循环节在有些题目中我们需要对指数进行降幂处理才能计算。比如计算 其中和 这里由于很大，所以需要进行降幂。那么实际上有如下降幂公式 有了上述公式，很多题目就可以得到解决。题目：http://acm.fzu.edu.cn/problem.php?pid=1759
2015-09-09 19:22:45
1277
原创 HDU 5167 Fibonacci（DFS暴力搜索）
FibonacciTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1905 Accepted Submission(s): 479Problem DescriptionFollowing is the
2015-09-07 22:05:54
464
原创 HDU 5154 Harry and Magical Computer（找环）
Harry and Magical ComputerTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1743 Accepted Submission(s): 684Problem Description
2015-08-31 21:25:59
396
原创 ZOJ 3818 Pretty Poem
Pretty PoemTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %lluSubmit Status Practice ZOJ 3818DescriptionPoetry is a form of literature that uses aesthetic
2015-08-30 21:17:58
412
原创 HDU 5147 Sequence II （线段树求前缀和、后缀和）
Problem DescriptionLong long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence. Pleas
2015-08-29 09:11:26
378
原创 HDU 4998 Rotate （几何变换——旋转）
RotateTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 815 Accepted Submission(s): 389Special JudgeProblem DescriptionNoting i
2015-08-28 15:15:38
554
原创 几何变换（二维、三维）
几何变换窗口区到视图区的坐标变换 实际的窗口区与视图区往往不一样大小，要在视图区正确地显示形体的，必须将其从窗口区变换到视图区。 比例关系，两者的变换公式为： 可以简单地将两者的关系表示为： 二维图形的几何变换 正如我们在附录中提到的那样，用齐次坐标表示点的变换将非常方便，因此在本节中所有的几何变换都将采用齐次坐标进
2015-08-28 13:40:35
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原创 HDU 5139 Formula（递推公式、离线处理）
FormulaTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1094 Accepted Submission(s): 384Problem Descriptionf(n)=(∏i=1nin−i+1)%
2015-08-27 09:51:40
432
原创 HDU 5144 NPY and shot（三分角度）
NPY and shotTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 761 Accepted Submission(s): 309Problem DescriptionNPY is going to
2015-08-27 09:35:07
438
原创 HDU 5143 NPY and arithmetic progression （枚举基本类型123、234、1234的个数）
NPY and arithmetic progressionTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 799 Accepted Submission(s): 254Problem Description
2015-08-27 08:50:13
605
原创 HDU 5407 CRB and Candies
CRB and CandiesTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 584 Accepted Submission(s): 292Problem DescriptionCRB has N di
2015-08-26 15:27:06
463
原创 HDU 5419 Victor and Toys （）
Victor and ToysTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 467 Accepted Submission(s): 157Problem DescriptionVictor has
2015-08-25 10:59:39
383
原创 HDU 5056 Boring count（不超过k个字符的子串个数）
Boring countTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 828 Accepted Submission(s): 342Problem DescriptionYou are given a
2015-08-24 10:26:19
748
原创 莫比乌斯函数
莫比乌斯函数在讲这个函数之前。最好先了解欧拉函数。我们用 \ 记为整除。 记得小学的时候整除和整除以的概念么？别混淆。 2整除4 记作 2\4。欧拉函数用来表示。那么根据法里级数的展开(这个感觉和ACM关系不大就先不介绍了。大概讲的就是构造所有最简分数的一种树。而法里级数n定义分母比如对于分母为12.化简后:分别
2015-08-21 16:00:57
725
原创 HDU 5382 GCD?LCM! （递推公式 + 打表）
GCD?LCM!Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 192 Accepted Submission(s): 105Problem DescriptionFirst we define:
2015-08-19 18:21:54
1072
原创 HDU 5396 Expression（ 组合+dp ？？）
ExpressionTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 380 Accepted Submission(s): 214Problem DescriptionTeacher Mai has n
2015-08-19 16:09:08
309
原创 HDU 5400 Arithmetic Sequence（左d1等差右d2等差的区间个数）（枚举位置，求最大区间，算区间个数）
Arithmetic SequenceTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 636 Accepted Submission(s): 285Problem DescriptionA sequen
2015-08-19 14:58:01
902
原创 HDU 5402 Travelling Salesman Problem （模拟 有规律）（左上角到右下角路径权值最大，输出路径）
Travelling Salesman ProblemTime Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 568 Accepted Submission(s): 200Special JudgeProblem De
2015-08-19 10:44:41
921
原创 HDU 5399 Too Simple （区分映射与函数）（）
Too SimpleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 449 Accepted Submission(s): 158HDU 5399 Too Simple （）Problem DescriptionRhason Cheung had
2015-08-18 21:14:26
585
原创 HDU 5101 Select（不同集合的两个数大于k的方案数）
SelectTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1422 Accepted Submission(s): 395Problem DescriptionOne day, Dudu, the m
2015-08-18 11:40:43
921
原创 HDU 5100 Chessboard （用k×1的矩形覆盖n×n的正方形）（找规律）
用 k × 1 的小矩形覆盖一个 n × n 的正方形棋盘，往往不能实现完全覆盖（比如，有时候 n × n 甚至根本就不是 k 的整倍数）。不过，在众多覆盖方案中，总有一种覆盖方案会让没有覆盖到的方格个数达到最少，我们就用 m(n, k) 来表示这个数目。求证：不管 n 和 k 是多少， m(n, k) 一定是一个完全平方数。如果 n 2 ，这是一个完全平方数。下面我们就只
2015-08-18 09:54:38
1028
原创 HDU 5087 Revenge of LIS II（次大递增子序列）
Revenge of LIS IITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1258 Accepted Submission(s): 423Problem DescriptionIn comput
2015-08-17 16:03:11
591
2
原创 HDU 5088 Revenge of Nim II （高斯消元？？）
Revenge of Nim IITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 589 Accepted Submission(s): 208Problem DescriptionNim is a m
2015-08-17 15:50:22
417
原创 Lucas定理（C（n,m）mod p）
Lucas定理卢卡斯定理及证明：表达式也可以表示为：Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)简单的说，Lucas定理是用来求C(n,m) mod p的值（p是素数）。#include #include using namespace std;#define MAX 150000long long fac[MAX];
2015-08-15 16:25:28
984
原创 HDU 5084 HeHe （找规律）
HeHeTime Limit: 5000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 311 Accepted Submission(s): 116Problem DescriptionM=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜tn−1tn
2015-08-15 10:55:09
577
原创 HDU 5083 Instruction（大模拟）
InstructionTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 928 Accepted Submission(s): 217Problem DescriptionNowadays, Jim Gr
2015-08-15 09:50:29
561
原创 HDU 5366 The mook jong （排列组合 或 找规律）
The mook jongAccepts: 506 Submissions: 1281 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Problem DescriptionZJiaQ want to becom
2015-08-14 17:15:46
571
原创 HDU 5365 Run （简单分析+暴力枚举）
RunTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionAFA is a girl who like runing
2015-08-14 16:13:15
436
原创 HDU 5387 Clock （简单计算）
ClockTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionGive a time.(hh:mm:ss)，you
2015-08-14 11:17:38
501
原创 HDU 5373 The shortest problem （简单模拟）
The shortest problemTime Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionIn this prob
2015-08-14 09:41:35
356
原创 HDU 5371 Hotaru's problem（Manacher算法）
Hotaru's problemTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s):0 Accepted Submission(s): 0Problem DescriptionHotaru Ichijou r
2015-08-14 09:24:57
323
原创 HDU 5067 Harry And Dig Machine (DFS)
Harry And Dig MachineTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 784 Accepted Submission(s): 302Problem Description As w
2015-08-13 09:53:38
337
原创 回文字符串—— Manacher算法
一、问题描述现给定一个已知的字符串str[]，现在想要在O(n)的时间复杂度之内求出一个最长的回文子字符串(正着和倒着顺序读一致)。Manacher最早发现了可以用O(n)的时间复杂度来解决该问题，所以这种方法称之为Manacher算法。二、符号说明回文串包括奇数长的和偶数长的，一般求的时候都要分情况讨论，Manacher的这个算法做了个简单的处理，把奇偶情况统
2015-08-11 21:00:18
402
原创 HDU 4686 Arc of Dream（矩阵快速幂）
Arc of DreamTime Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2992 Accepted Submission(s): 921Problem DescriptionAn Arc of Drea
2015-08-07 18:53:27
380
原创 HDU 5358 First One（枚举）
First OneTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 690 Accepted Submission(s): 205Problem Descriptionsoda has an inte
2015-08-07 11:46:49
523
原创 HDU 5020 Revenge of Collinearity（枚举）
Revenge of CollinearityTime Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Problem DescriptionIn geometry, collinearity is a property of a set of points, s
2015-08-06 21:36:08
741
1
原创 HDU 5019 Revenge of GCD（简单枚举）
Revenge of GCDTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Problem DescriptionIn mathematics, the greatest common divisor (gcd), also known as the g
2015-08-06 20:13:07
605
原创 HDU 5015 233 Matrix （数论——矩阵快速幂）
233 MatrixTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionIn our daily life we
2015-08-05 19:15:26
677
原创 HDU 5351 MZL's Border（找规律+高精度）
MZL's BorderTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 895 Accepted Submission(s): 287Problem DescriptionAs is known to
2015-08-05 18:49:13
518
原创 HDU 5344 MZL's xor （异或运算的性质）
MZL's xorTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionMZL loves xor very muc
2015-08-04 20:07:02
719
罗塞塔石碑安卓版
2018-06-09
高斯消去法求解线性方程组C/C++程序（输入输出支持分数和浮点数）
2018-06-09
遗传算法与最小二乘法拟合函数比较的C/C++代码实现
2018-06-09
USB转串口驱动（串口芯片的型号PL2303）（支持win8/8.1系统）
2016-07-12
USB转串口驱动（串口芯片的型号PL2303）
2016-07-12
51单片机C语言60个复杂程序
2016-07-04
51单片机50个相对复杂一些的程序
2016-07-04
中国象棋的LABVIEW编程
2016-07-04
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