package com.ywx.count; import java.util.Scanner; /** * @author Vashon * date:20150410 * 题目:计算1至n的k次方的和 * */ public class Sumdemo03 { public static void main(String[] args) { System.out.println("请输入两个数(计算1至n的k次方的和):"); Scanner sc1=new Scanner(Sys

We are given a binary tree (with root node root), a target node, and an integer value K. Return a list of the values of all nodes that have a distance K from the target node.  The answer can be returned in any order. Example 1: Input: root = [3,5,1,6

[本文链接] http://www.cnblogs.com/hellogiser/p/find-k-missing-numbers-from-1-to-n.html  [题目] 从1到n,共有n个数字(无序排列),每个数字只出现一次.现在随机拿走一个数字x,请给出最快的方法,找到这个数字.要求时间复杂度为O(n),空间复杂度为O(1).如果随机拿走k(k>=2)个数字呢? [分析] 题目给出的条件很强,数字是从1~n的数字,限制了数字的范围:每个数字只出现一次,限制了数字出现的次数:随即拿走了一

Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible

/** 题目:hdu6003 Problem Buyer 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6003 题意:给定n个区间,以及m个数,求从n个区间中任意选k个区间,满足m个数都能在k个区间中找到一个包含它的区间,如果一个区间包含了x,那么 该区间不能再去包含另一个数,即k>=m.求最小的k.如果不存在这样的k,输出“IMPOSSIBLE!”. 思路:贪心; 对n个区间,先左端点进行由小到大排序,然后右端点由大到小.对m个数由小到大排序. 对

1.题目:输入n个整数,找出其中最小的K个数. 例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4. 快排思路(掌握): class Solution { public: vector<int> GetLeastNumbers_Solution(vector<int> input, int k) { vector<int> result; || input.size()<k) return result; ; ; int index

#include<bits/stdc++.h> using namespace std; #define ll long long #define maxn 100010 /* 已知一个序列A1.A2….An,给你一个整数K,找到满足所有Ai+Aj>=k的数对(i,j)的个数 */ int main() { ],k,low,upp; while(cin >> n >> k){ , ans2 = ; ; i<n; i++) cin >> a[i];

对于曾经,假设要我求第k小元素.或者是求前k大元素,我可能会将元素先排序,然后就直接求出来了,可是如今有了更好的思路. 一.线性时间内求第k小元素 这个算法又是一个基于分治思想的算法. 其详细的分治思路例如以下: 1.分解:将A[p,r]分解成A[p,q-1]和A[q+1,r]两部分.使得A[p,q-1]都小于A[q],A[q+1,r]都不小于A[q]; 2.求解:假设A[q]恰好是第k小元素直接返回,假设第k小元素落在前半区间就到A[p,q-1]递归查找.否则到A[q+1,r]中递归查找. 3

题目描述:给定K个字符数组,从这k个字符数组中任意取一个字符串,按顺序拼接,列出所有可能的字符串组合结果! 样例: input:[["a","b","cd",["de"],["e","f"]] output:["adee","adef","bdee","bdef","cddee",&quo

Nlogonia is fighting a ruthless war against the neighboring country of Cubiconia. The Chief General of Nlogonia's Army decided to attack the enemy with a linear formation of soldiers, that would advance together until conquering the neighboring count

Given a sorted array, two integers k and x, find the kclosest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred. Example 1: Input: [1,2,3,4,5], k=4, x=3 Outpu

题目:Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题意:把k个排序成一个有序链表. 用优先队列先把k个链表遍历一遍把值存起来,在建一个新链表吧数从优先队列里一个个放进去,注意空指针的判断. /** * Definition for singly-linked list. * struct ListNode { * int val; * List

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2  Note: The length of the array is in range [1, 20,000]. The range of number

这里思路是堆排序,而且是小根堆.C++中包含在头文件<queue>的priority_queue本质就是堆排序实现的.其中priority_queue函数原型是 priority_queue<Type, Container, Functional>         Type 就是数据类型,        Container 就是容器类型(Container必须是用数组实现的容器,比如vector,deque等等,但不能用 list.STL里面默认用的是vector),       

我们知道K/3 Cloud和云之家进行集成,在管理中心里面有个设置,移动账套启用,只能支持一个账套启用那么能不能支持两个账套部署到云之家中呢?其实移动BOS平台默认是支持,答案就在发布到云之家的菜单中:下面我们来看一下,我们发布到云之家的部署菜单一般是什么样子的  http://XXX/K3Cloud/mobile/k3cloud.html?entryrole=XT&appid=10037&formId=MOB_WFTodoList&formType=mobilelist好,那我们

g(i)=k*i+b; 0<=i<nf(0)=0f(1)=1f(n)=f(n-1)+f(n-2) (n>=2)求f(b) +f(k+b) +f(2*k+b) +f((n-1)*k +b) 之和 Sample Input2 1 4 100 // k b n MOD2 0 4 100 Sample Output2112 矩阵A      相当于 1 1          f(2)  f(1) 1 0          f(1)  f(0) | 1       1| ^b          |

Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred. Example 1: Input: [1,2,3,4,5], k=4, x=3 Outp

Given a non-empty array of integers, return the k most frequent elements. For example,Given [1,1,1,2,2,3] and k = 2, return [1,2]. Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements. Your algorithm's time complexity must be bet

k个数乘(cheng) 题目描述 桐桐想把一个自然数N分解成K个大于l的自然数相乘的形式,要求这K个数按从小到大排列,而且除了第K个数之外,前面(K-l)个数是N分解出来的最小自然数.例如:N=24,K=2时,输出为24=2×12,而不是24=4×6:如N=3,K=I时,3=3; N=3,K=2时,输出则为"No answer!".你能帮助她吗? 输入 第1行:N(2≤N≤107): 第2行:K(1≤K≤100). 输出 输出样例格式的分解式. 样例输入 24 2 样例输出 24=2*

/* * POJ-1012.cpp * * Created on: 2013年10月31日 * Author: Administrator */ #include <iostream> #include <cstdio> using namespace std; int x[] = { 2, 7, 5, 30, 169, 441, 1872, 7632, 1740, 93313, 459901, 1358657, 2504881 }; int main(){ int n; whil

// test20.cpp : 定义控制台应用程序的入口点. // #include "stdafx.h" #include<iostream> #include<vector> #include<string> #include<queue> #include<stack> #include<cstring> #include<string.h> #include<deque> #incl