#include <iostream> #include <algorithm> #include <vector> int main() { int n, r; n = ; r = ; std::vector<bool> v(n); std::fill(v.end() - r, v.end(), true); std::vector< std::vector<int> > sequence_vector; do { std::vec…

排列(Arrangement),简单讲是从N个不同元素中取出M个,按照一定顺序排成一列,通常用A(M,N)表示.当M=N时,称为全排列(Permutation).从数学角度讲,全排列的个数A(N,N)=(N)*(N-1)*...*2*1=N!,但从编程角度,如何获取所有排列?那么就必须按照某种顺序逐个获得下一个排列,通常按照升序顺序(字典序)获得下一个排列. 例如对于一个集合A={1,2,3,},首先获取全排列a1: 1,2,3,:然后获取下一个排列a2: 1,3,2,:按此顺序,A的全排列如下…

平常需要全排列的时候,一般都是dfs然后字符串匹配啥的……今天看题解的时候突然发现了这个神器. next_permutation()函数在c++的algorithm库里,作用是传入一个数组,输出这个数组的后一个排序.同样还有prev_permutation()输出前一个排序.这个“后一个”指的是,不存在另一个排序方案,字典序大于目前排序而小于将要输出的排序.(换句话说,把它的左右排列按字典序升序排序,输出当前状态的下一个) next_permutation()是一个bool函数,当前序列不存在下…

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target. Example: nums = [1, 2, 3] target = 4 The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1,…

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Ensure that numbers within the set are sorted in ascending order. Example 1…

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination. Note: All numbers (including target) will be posi…

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times. Note: All numbers (including target) will…

[本文链接] http://www.cnblogs.com/hellogiser/p/string-combination.html [题目] 题目:输入一个字符串,输出该字符串中字符的所有组合.举个例子,如果输入abc,它的组合有a.b.c.ab.ac.bc.abc. [分析] 在之前的博文28.字符串的排列[StringPermutation]中讨论了如何用递归的思路求字符串的排列.同样,本题也可以用递归的思路来求字符串的组合. [递归法求组合] 可以考虑求长度为n的字符串中m个字符的组合,…

问题 Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target. Example: nums = [1, 2, 3] target = 4 The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (…

这是一个c++函数,包含在头文件<algorithm>里面,下面是基本格式. 1 int a[]; 2 do{ 3 4 }while(next_permutation(a,a+n)); 下面的代码可产生1~n的全排列. #include <stdio.h> #include <algorithm> using namespace std; int main(){ int n; while(scanf("%d",&n)&&n){…