[题目链接] https://codeforces.com/contest/204/problem/E [算法] 首先构建广义后缀自动机 对于自动机上的每个节点 , 维护一棵平衡树存储所有它所匹配的字符串编号 可以通过启发式合并得到 计算答案时 , 我们枚举每个右端点 , 当当前集合大小 < K时 , 不断走向link节点 时间复杂度 : O(NlogN) [代码] #include<bits/stdc++.h> using namespace std; typedef long lon…

传送门在这里. 大意: 给一堆字符串,询问每个字符串有多少子串在所有字符串中出现K次以上. 解题思路: 这种子串问题一定要见后缀自动机Parent树Dfs序统计出现次数都是套路了吧. 这道题统计子串个数,那么可以发现,若一个节点所对应的子串出现了K次,那么其贡献就是len,不需要考虑重复. 因为即使出现重复也是在两个位置. 那么只需统计以每个点结束的子串就好了. 之前的Dfs序就很套路了. 只需再跑一遍字符串,更新答案就好了. 代码: #include<cstdio> #include<…

E. Little Elephant and Strings time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant loves strings very much. He has an array a from n strings, consisting of lowercase Englis…

CodeForces - 204C Little Elephant and Furik and Rubik 个人感觉是很好的一道题 这道题乍一看我们无从下手,那我们就先想想怎么打暴力 暴力还不简单?枚举所有字串,再枚举所有位置,算出所有答案不就行了 我们自然不能无脑暴力,但是暴力可以给我们启发 我们知道所有对答案做出贡献的字符一定是相同的(废话) 所以我们可以O(n^2)首先枚举两个字符串中相同的字符然后再考虑如何贡献 然后计算出所有的方案下的值,再除以n*(n+1)*(2*n+1)/6 [不知…

The Little Elephant loves strings very much. He has an array a from n strings, consisting of lowercase English letters. Let's number the elements of the array from 1 to n, then let's denote the element number i as ai. For each string ai (1 ≤ i ≤ n) t…

题目描述: Little Elephant and Interval time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant very much loves sums on intervals. This time he has a pair of integers l and r (l ≤ r…

Codeforces 204 E 题意:给\(n\)个串,求对于每一个串在至少\(k\)个串中出现的它的子串\(S_{l..r}\)有多少个. 思路:后缀自动机上\(dp\)... 我们首先构造出这\(n\)个串的后缀自动机,其中需要注意将某个串的构建完了后直接将\(lst\)指针赋为\(root\),那么就可以包含这些串的所有子串并且不会有问题. 然后我们就考虑如何来算出某一个子串在\(n\)个串中出现了多少次. 假设现在我们从\(S_i\)的开头走到第\(j\)位,走到了节点\(u\),那么…

题意:有n个串,询问每个串有多少子串在n个串中出现了至少k次. 题解:sam,每个节点开一个set维护该节点的字符串有哪几个串,启发式合并set,然后在sam上走一遍该串,对于每个可行的串,所有的fail都是可行的直接加上len,不可行就往fail上跳. 被for(int i=0;s[i];i++)给骚死了,一直tle.... //#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#prag…

 Little Elephant and Chess Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 259A Description The Little Elephant loves chess very much. One day the Little Elephant and his friend decided…

C. Simple Strings 题目连接: http://www.codeforces.com/contest/665/problem/C Description zscoder loves simple strings! A string t is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add a…

Little Elephant and Array Time Limit: 4000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 221D64-bit integer IO format: %I64d      Java class name: (Any) The Little Elephant loves playing with arrays. He has array a,…

题目链接:http://codeforces.com/contest/682/problem/D 给你两个字符串,求两个字符串中顺序k个的相同子串 长度之和.(注意是子串) dp[i][j][k][0] 表示a[i] == a[j]时,a字符串前i个和b字符串前j个,顺序k个相同的子串 长度之和 dp[i][j][k][1] 表示a[i] != a[j]时,顺序k个相同子串的长度之和 dp[i][j][k][0] = max(dp[i - 1][j - 1][k][0], dp[i - 1][j…

http://codeforces.com/problemset/problem/204/A 题意:给定一个[L,R]区间,求这个区间里面首位和末尾相同的数字有多少个 思路:考虑这个问题满足区间加减,我们只考虑[1,n],考虑位数小于n的位数的时候,我们枚举头尾的数是多少,然后乘上10的某幂次,再考虑位数相等时,从高位往低位走,先考虑头尾数字小于最高位的情况,也像刚才那个随便取,当头尾数字等于最高位时,从高往低走,先算不与这位相等的,走下一步就代表与这位相等.最后要记得判断一下如果原数字头等于尾…

Description You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new…

Little Elephant and Broken Sorting 怎么感觉这个状态好难想到啊.. dp[ i ][ j ]表示第 i 个数字比第 j 个数字大的概率.转移好像比较显然. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair…

time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output: standard output Given a string sss of length nnn and integer k (1≤k≤n)k\ (1≤k≤n)k (1≤k≤n). The string sss has a level xxx, if xxx is largest non-negative…

Game with Strings 题意并不是在图上走,看了好久才看出来.. dp[ i ][ mask ]表示从 i 层开始走,起点有mask个, a的个数-b的个数的  最大值或者最小值. #include<bits/stdc++.h> #define LL long long #define LD long double #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL&g…

B - DZY Loves Strings DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter c DZY knows its value wc. For each special string s = s1s2... s|s|(|s| is the length of the string) he represents its value wit…

Little Elephant and LCM #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()…

D. Equivalent Strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal l…

C. Two strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another)…

[题目]D. Picking Strings [题意]给定只含'A','B','C'的字符串,支持以下变换:1.A - BC   2.B - AC   3.C - AB   4.AAA - empty string(左边变成右边) 给定S串和T串,q次询问,每次给出S串的一个子串x和T串的一个子串y,求x是否能变到y.n,m,q<=10^5. [算法]模拟??? [题解]观察一些规律,首先B和C等价:B-AC-AAB-AAAC-C. 然后B前面的A可以消除:AB-AAC-AAAB-B. 所以新的…

Discription The Little Elephant loves permutations of integers from 1 to n very much. But most of all he loves sorting them. To sort a permutation, the Little Elephant repeatedly swaps some elements. As a result, he must receive a permutation 1, 2, 3…

C. Dreamoon and Strings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a…

题目链接: D. Alyona and Strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After returned from forest, Alyona started reading a book. She noticed strings s and t, lengths of which are n a…

先上题目: Little Elephant and Sorting time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The Little Elephant loves sortings. He has an array a consisting of n integers. Let's number the array elem…

[题目链接]:http://codeforces.com/contest/798/problem/B [题意] 给你n个字符串; 每次操作,你可以把字符串的每个元素整体左移(最左边那个字符跑到最后面去了) 问你最少经过多少次操作可以使得所有字符串都相同; [题解] 枚举最后每个字符串都变成了哪一个字符串; 然后每个字符串都模拟一下左移的过程;直到相等记录总的移动次数; 或者左移超过了长度的次数;输出不可能能和目标串一样; 记录最小的移动次数就好; [Number Of WA] 0 [完整代码]…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output After returned from forest, Alyona started reading a book. She noticed strings s and t, lengths of which are n and m respectively. As usual, rea…

CF258D Little Elephant and Broken Sorting 题意 题意翻译 有一个\(1\sim n\)的排列,会进行\(m\)次操作,操作为交换\(a,b\).每次操作都有\(50\%\)的概率进行. 求进行\(m\)次操作以后的期望逆序对个数. \(n,m\le 1000\) 输入输出格式 输入格式: The first line contains two integers \(n\) and \(m\) \((1\leq n,m\leq 1000,n>1)\) -…

传送门:https://codeforces.com/problemset/problem/220/B 题意: 给你n个数,m次询问,每次询问问你在区间l,r内有多少个数满足其值为其出现的次数 题解: 莫队算法 思路:每次区间向外扩的更新的过程中,检查该位置的数ai的出现次数是否已经达到ai或ai+1,以判断是否要更新结果.同理,区间收缩的时候判断ai出现次数是否达到ai或ai-1. 因为数据范围为1e9,所以需要离散化,方便记录每种数出现的次数 代码: /** * ┏┓ ┏┓ * ┏┛┗━━━…