Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false. Note: You may assume the string contains…

Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. Note:You may assume the string contains onl…

Problem: Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. Summary: 给出字符串s和t,判断t是不是s的相同字母异序词.…

http://poj.org/problem?id=3207 题意:一个圆上顺时针依次排列着标号为1-n的点,这些点之间共有m条边相连,每两个点只能在圆内或者圆外连边.问是否存在这些边不相交的方案.(n<=1000, m<=500) #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using n…

Valid Anagram My Submissions Question Total Accepted: 43694 Total Submissions: 111615 Difficulty: Easy Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return t…

class Solution(object):    def isAnagram(self, s, t):        if sorted(list(s.lower()))==sorted(list(t.lower())):            return True        return Falsesol=Solution()print sol.isAnagram(s='anAgram', t='nagaram')…

Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. Note:You may assume the string contains onl…

Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. 解法:英文颠倒词,只要两个单词中,英文字母出现的次数相等即可. 代码如下: publi…

字符串Hash+可持久化线段树 好神奇的转化…… 蒟蒻一开始还去想AC自动机去了……然而由于a[i]的范围是小于等于n,怎么也想不出一个时间复杂度合理的方法 膜拜了题解0.0原来是字符串Hash! 首先每个询问的长度都是固定的!K!那么我们就可以看作一个长度为K的数!然后就可以利用字符串Hash将一个长度为n的数列转化成一个长度为n-k+1的数列!查询一个特定的序列在一个区间中是否出现过,就变成了在一段区间中,某个数是否出现过!然后就可以用可持久化线段树来维护了…… 长度为K这个条件好神啊……居…