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poj3195
- poj3195 输入2个正整数，求出他们的最大公约数。 http://poj.grids.cn/problem?id=3195-poj3195 input two positive integers, find their common denominator. http://poj.grids.cn/problem?id=3195
1001
- POJ report of 1001 it s good for people who study acm
Poj1022
- 本题写了一个EE3的产品描述，是四维的。确实很难想像清楚。我们这样想就行。它有四个坐标，每一维都可以放上一个立方体，边长为1.然后就是怎么判断是否Inconsistent，有两种情况：1），如题上举的例子，ID为1的在x2坐标上前面是3，那么ID为3的在X2坐标的后面应该是1，否则就Inconsistent。-We usually think that there are three geometric dimensions the fourth dimension is usually ti
1904
- POJ 1904 King s Quest Time Limit:15000MS Memory Limit:65536K-King s Quest Time Limit:15000MS Memory Limit:65536K Total Submit:466 Accepted:130 Case Time Limit:2000MS Descr iption Once upon a time there lived a king and he had N
Poj_1011_sticks
- 由题意知道，本题就是根据题目给出的木棒，拼出长度最短的等长的棒子，本题主要应用了深度优先搜索+强剪枝。我们知道，肯定是先把小棒排序，从最大的开始枚举，然后如果到某一个长度时，满足条件，哪么就输出该长度。-George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the origi
Poj_1018_CommunicationSysterm
- 本题一看一位恨水，没想到还有这么多细节啊！吸取教训。仔细一看想用dp，找不到什么子结构，贪心感觉有不太对。最后感觉有点像二分的枚举。题目意思大概就是有n个devices，每个都需要一个bandwidth，而这个会有很多商家提供，价格不同，现在在n个中，每个选一个，但是着n个中选B最小的，而所有选的price和最小，即是B/p最大。-We have received an order from Pizoor Communications Inc. for a special communicati
Poj_2796
- 本题主要就是要求一个数列里面，连续的几个数的和乘以这个连续段里最小的一个数，在所有这样连续数列里最大的一个，直接输出最大值就可以了。-Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people s memories about some period
poj1469
- poj1469, 二分图最大匹配，最大匹配-poj1469, Maximum Bipartite Matching
2352
- 线段树和树状数组的运用，ACM知识，该题为poj 上的2352题的详细代码-An array of trees and tree line use, ACM knowledge, which entitled the 2352 title poj detailed code
2777
- 线段树和树状数组的运用，ACM知识，该题为poj 上的2777题的详细代码-An array of trees and tree line use, ACM knowledge, which entitled the 2777 title poj detailed code
2586
- poj.pku.edu.cn 北大poj 2586 C++代码,已提交-for the 2586 problem on pku.edu.cn
4703829_AC_0MS_400K
- POJ ACM CODE _AC_0MS_400K.-POJ ACM CODE _AC_0MS_400K.cc
Programming_guide_and_online_practice-4.23bylwx.ra
- POJ各类问题讲解。适合初学者。书上的练习题都很有代表性-POJ on various issues. Suitable for beginners. Exercises of the book are representative
5627957_AC_16MS_128K
- POJ 3295 纯C AC代码 方法极简单~-POJ 3295 Methods of pure C AC code simple ~
POJ
- PKU 1000-1050我ac的代码！绝对保证质量！-PKU 1000-1050 code I ac! Absolute guarantee of quality!
1051
- poj 1051 字符串处理，简单题但是要注意细节！-poj 1051 string handling, a simple question but attention to detail!
Beginning
- poj的几道入门题，一定要自己亲手做。呵呵，不起跬步无以至千里！-poj entry of several questions, we must do their own hands. Oh, can not afford even brief step-by-step journey of a thousand miles-free!
3273
- poj 3273 二分解法，逐位判断是否满足条件-poj 3273
3720
- poj 3720多个版本的ac代码 适合算法爱好者和acmer-poj 3720 multiple versions of code for algorithm ac fan and acmer
3726
- poj 3726多个版本的ac代码 适合算法爱好者和acmer-poj 3726 multiple versions of code for algorithm ac fan and acmer