algorithm
POJ1961
POJ1961 Period Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 15006 Accepted: 7158 Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.
POJ1686
POJ1686 Lazy Math Instructor Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 3766 Accepted: 1307 Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading very hard. So, the instructor needs help from computer programmers and you can help.
Simple implementation of ls command
Just an assignment from my Linux programming course. I have only done a small part of work of the ls command, after all, it’s a program with 400+ lines, if you want to study full source codes of ls command(in which there are 4000+ lines codes), you can get it from here.
UVAOJ11198
UVA11198 Rujia Liu的神题啊，虽然AC率有40%，但是AC的人数只有两三百个。。随便搞一搞就前150了 Orz。。。 题意就是给一个序列，只有当两个元素异号且绝对值之和是质数才可以移动到彼此的左右边，即ai可以移动到aj的左边或者右边，aj也可以移动到ai的左边或者右边，问最少移动多少次就可以使整个序列递增排列。
Codeforces #527 C
527C C. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular wmm × h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
UVAOJ10001
UVAOJ10001 Great thanks to: 孟海 难点不在算法，在题意。 题目给一个字符串，然后求是否有一个字符串能够通过题目所给的自动机规则变换过来，如果存在，则输出REACHABLE，否则输出GARDEN OF EDEN，就是寓意始祖，即这个字符串无法通过所给的自动及变换规则变换过来。
AOAPC I: Volume 3.Brute Force-Backtracking:Easy
<<算法竞赛入门经典>> UVaoj第四卷暴力求解简单回溯习题: 10474-Where is the Marble? 之前的习题卷里面也收录了这个习题，输入N个数, 然后再输入要查找的Q个数, 依次输出Q个数中每个数在N个数中被找到的位置, 找不到就not found.