UVAlive 3026 Period 题目: Period   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive…

UVAlive 3135 Argus Argus Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, fin…

UVAlive 3942 Remember the Word 题目: Remember the Word   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Neal is very curious about combinatorial problems, and now here comes a problem about words. Kn…

UVAlive 4329 Ping pong 题目: Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player ha…

UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent…

UVAlive X-Plosives 思路:    “如果车上存在k个简单化合物,正好包含k种元素,那么他们将组成一个易爆的混合物”  如果将(a,b)看作一条边那么题意就是不能出现环,很容易联想到Kruskal算法中并查集的判环功能(新加入的边必须属于不同的两个集合否则出现环),因此本题可以用并查集实现.模拟装车过程即可. 代码: #include<cstdio> #include<cstring> #define FOR(a,b,c) for(int a=(b);a<(c…

UVAlive 4670 Dominating Patterns 题目:   Dominating Patterns   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description The archaeologists are going to decipher a very mysterious ``language". Now, they kno…

题意:给你一个字符串,问第i位前是否有循环节,若存在,则循环节是多少? 思路:考察失配函数f[i]的意义.只要i%(i-f[i])==0,则循环节长度为i/(i-f[i]).字符在[0,f[i]],[i-f[i],i]范围内的相等,所以如果存在循环节则每i-f[i]可以分为一段.理解起来比较抽象,模拟一遍. #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #in…

input n 2<=n<=1000000 长度为n的字符串,只含小写字母 output Test case #cas 长度为i时的最小循环串 循环次数(>1) 若没有则不输出 做法:若next数组是连续的整数且next[i]+1是错位部分长度的倍数,则必有循环出现,且循环次数为(i+1)/错位部分长度 #include <cstdio> #include <queue> #include <cstring> #include <iostream…

kmp的代码很短,但是不太容易理解,还是先说明一下这个算法过程吧. 朴素的字符串匹配大家都懂,但是效率不高,原因在哪里? 匹配过程没有充分利用已经匹配好的模版的信息,比如说, i是文本串当前字符的下标,j是要匹配的模版串当前正在匹配的字符的下标.(下标都从零开始,j同时可以表示已经匹配的字符长度) 当匹配到i = 4, j = 4的时候失配了,朴素的匹配做法是往右边移一位然后从j开始扫,这样做效率很低. 不难发现前面已经匹配好的串ab是最长公共前缀后缀.把串移动到后缀的第一个位置正好是 朴素的匹…