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…

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…