A typical game theory problem - Recursion + Memorized Searchhttp://justprogrammng.blogspot.com/2012/06/interviewstreet-challenges-permutation.html#.VlEMrvmrTIU #include<iostream> #include<set> using namespace std; // Bit Ops Utils int set(int…

The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s = "a…

Given a string, determine if a permutation of the string could form a palindrome. For example,"code" -> False, "aab" -> True, "carerac" -> True. Hint: Consider the palindromes of odd vs even length. What difference d…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replaceme…

The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

UVA - 11525 Permutation 题意:输出1~n的所有排列,字典序大小第∑k1Si∗(K−i)!个 学了好多知识 1.康托展开 X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[1]*0! 其中a[i]为第i位是i往右中的数里 第几大的-1(比他小的有几个). 其实直接想也可以,有点类似数位DP的思想,a[n]*(n-1)!也就是a[n]个n-1的全排列,都比他小 一些例子 http://www.cnblogs.com/hxsyl…

For research purpose, I've read a lot materials on permutation test issue. Here is a summary. Should be useful. Still, thanks for contributors online. P value calculation Because the actual value is one of those permutations, I would like to change t…

题目传送门:https://www.hackerrank.com/challenges/unique-colors 感谢hzq大神找来的这道题. 考虑点分治(毕竟是路经统计),对于每一个颜色,它的贡献是独立的.我们可以在一次点分治中合在一块处理(为什么时间复杂度是对的呢,因为我们每次改动只会根据当前点的颜色进行变动,而不是所有颜色对着它都来一遍).每次先对重心单独计算答案贡献,此时也将当前区域的各个答案贡献计算出来,并以此为基础(之后称之为基准贡献,即代码中的tot).对于每一棵子树,我们先df…