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()…

题意: input : n a1,a2,...,an 1 <= n <= 10^5 1 <= ai <= 10^5 求b数组的方案数,b数组满足: 1. 1 <= bi <= ai 2. lcm(bi) = max(bi) solution: 2说明了b数组的每一个元素都是max(bi)的约数 则: 1.sort a数组,ma  = a[n] 2.预处理2个数组,vector<int> dive[MAXN],int pos[MAXN] dive[i]保存i的…

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…

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…

 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…

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,…

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…

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…

[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…

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…

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…

A /*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; ][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }}; , gakki = + + + + 1e9; , MAXM = 2e…

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: /** * ┏┓ ┏┓ * ┏┛┗━━━…

ACM思维题训练集合 The Little Elephant has two permutations a and b of length n, consisting of numbers from 1 to n, inclusive. Let's denote the i-th (1 ≤ i ≤ n) element of the permutation a as ai, the j-th (1 ≤ j ≤ n) element of the permutation b - as bj. Th…

ACM思维题训练集合 The Little Elephant has got a problem - somebody has been touching his sorted by non-decreasing array a of length n and possibly swapped some elements of the array. The Little Elephant doesn't want to call the police until he understands i…

Codeforces 题目传送门 & 洛谷题目传送门 yyq:"hot tea 不常有,做过了就不能再错过了" 似乎这是半年前某场 hb 模拟赛的 T2?当时 ycx.ymx.yxh 都去打了,可我似乎那时候还在旅游? 不难发现,对于操作 \(i\) 实际上是将 \(a_i\) 与 \(b_i\) 子树的并集内的点的答案集合并上 \(a_i\) 与 \(b_i\) 子树的并集.于是我们考虑将询问离线下来并在树上进行一遍 DFS,动态地维护一个可重集 \(S\).当我们第一次访问…

This problem can be solve in simpler O(NsqrtN) solution, but I will describe O(NlogN) one. We will solve this problem in offline. For each x (0 ≤ x < n) we should keep all the queries that end in x. Iterate that x from 0 to n - 1. Also we need to kee…

题意:N个数,M个查询,求[Li,Ri]区间内出现次数等于其数值大小的数的个数. 分析:用莫队处理离线问题是一种解决方案.但ai的范围可达到1e9,所以需要离散化预处理.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. 另一种更高效的方法是使用树状数组离线处理查询.用一个vector数组维护每个ai以此出现的位置.显然ai>N的数不会对结果做出贡献,所以数组开1e5就足够了.树…

题意: m次查询.每次查询范围[L,R]中出现次数等于该数字的数字个数. 题解: 由于分块,在每次询问中,同一块时l至多移动根号n,从一块到另一块也是最多2倍根号n.对于r,每个块中因为同一块是按y排序,所以最多移动根号n:一共根号n个块.注意l和r要分开考虑. 要注意的是这道题需要离散一下数据. #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #incl…

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<…

A. Little Elephant and Chess 模拟. B. Little Elephant and Magic Square 枚举左上角,计算其余两个位置的值,在\(3\times 3\)判断是否符合题意. C. Little Elephant and Bits 去掉最前面的0,可以使后面的1往前移. D. Little Elephant and Elections 对于选中的数,我们只关心其中的lucky number个数,容易想到用数位dp来求,结果用\(c_i\)表示有\(i\…

Little Elephant and Magic Square CodeForces - 259B Little Elephant loves magic squares very much. A magic square is a 3 × 3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table…

题目地址: 选题为入门的Codeforce div2/div1的C题和D题. 题解: A:CF思维联系–CodeForces -214C (拓扑排序+思维+贪心) B:CF–思维练习-- CodeForces - 215C - Crosses(思维题) C:CF–思维练习–CodeForces - 216C - Hiring Staff (思维+模拟) D:CF思维联系–CodeForces-217C C. Formurosa(这题鸽了) E:CF思维联系–CodeForces - 218C E…

[codeforces 235]A. LCM Challenge 试题描述 Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three…

A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't wa…

A. Elephant 题目连接: http://www.codeforces.com/contest/617/problem/A Descriptionww.co An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point x(x > 0) of the coord…

C. Little Elephant and Shifts Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/220/C Description The Little Elephant has two permutations a and b of length n, consisting of numbers from 1 to n, inclusive. Let's de…

C. The Big Race Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/592/problem/C Description Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of…

题目链接:http://codeforces.com/problemset/problem/1154/G 题目大意: 给定n个数,在这些数中选2个数,使这两个数的最小公倍数最小,输出这两个数的下标(如果有多组解,任意输出一组即可). 分析: 直接2个循环两两配对一下肯定要超时的,这里可以考虑枚举公因数,因为LCM是与最大公因数相关的,如果我们枚举了所有的公因数,那么我们就枚举了所有的LCM. 对于每一个公因数d,含有因子d的数从小到大排序为x1,x2,x3……xn,共有n个. 首先计算一下前两个…