目前HVIE里查询如下语句报错：select * from dw.ticket_user_mtime limit 10;错误如下：17/07/06 16:45:38 [main]: DEBUG impl.RecordReaderImpl: merge = [{data range [22733, 19927580), size: 19904847 type: array-backed}]Failed with exception java.io.IOException:java.lang.ArrayIndexOutOfBoundsException: 102417/07/06 16:45:38 [main]: ERROR CliDriver: Failed with exception java.io.IOException:java.lang.Array...
传送门
题目大意
你会得到一个数组a，并且会有q个查询，给定两个数字i和x，将a[i]乘以x，输出每次查询后的数组a的gcd
题解
牛客的寒假训练营有一道比这个简单的同类型题，这里上个牛客的传送门，我们先考虑如果数组a没有被修改，应该怎么求他们的gcd，很显然，我们可以将每个数字进行质因子分解，这样数组a中所有数都有的质因子，一定就是他们的gcd的质因子。再考虑如果有修改会怎么样，如果a[i]乘x，将x也质因子分解，如果分解出的...
Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).Return the running sum of nums. Example 1:Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].Constraints:1 <= nums.length <= 1000-10^6 <= nums[i] <= 10^6时间复杂度：O(n)空间复杂度：O(1)class Solution {publicint[] runningSum(int[] nums) ...
java.lang.Reflect.Array类提供了动态创建和访问数组元素的各种静态方法。package com.sunchao.reflection;import java.lang.reflect.Array;/*** The array reflection.* @author Administrator**/publicclass ArrayReflection {publicstaticvoid main(String args[]) throws Exception {Class<?> clazz = Class.forName("java.lang.String");String[] array = (String[]) Array.newInstance(clazz, 10);Array.set(array, 5, "hell...
仅供自己学习
思路：因为num1的空间能容纳num1和num2的内容，所以第一种方法就是可以把num2的数全加进num1中然后在sort即可。 相当于调用快排时间复杂度为O（（m+n）log（m+n）），空间复杂度O（1）
第二种方法是创建一个数组，然后比较num1和num2的大小放入新数组，最后再把新数组赋给num1. 时间复杂度O（m+n），空间复杂度O（m+n）上述方法比较简单就不加代码了。第三种方法就是原地算法，因为nums1的空间足够大，那么我们从最后...
题目连接http://acm.hdu.edu.cn/showproblem.php?pid=5280 Senior‘s ArrayDescriptionOne day, Xuejiejie gets an array $A$. Among all non-empty intervals of $A$, she wants to find the most beautiful one. She defines the beauty as the sum of the interval. The beauty of the interval---$[L,R]$ is calculated by this formula : beauty$(L,R) = A[L]+A[L+1]+……+A[R]$. The most beautiful interval is the one ...
一个根号分块经典题。
考虑如果我们直接\(dp\)求的话是\(O(n * n)\)的
我们发现这个dp的复杂度有一维是值域大小。
我们还发现我们其实没有必要把所有的答案都求出来，如果\(\sqrt n \leq k\)，那么最多不会超过\(\sqrt n\)步，我们完全可以进行暴力。
于是我们只需要处理\(k \leq \sqrt n\)的部分，dp的复杂度也变成\(O(n\sqrt n)\)了CF797E Array Queries
#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long l...
26. Remove Duplicates from Sorted Array
Easy
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Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Clarification:
Confused why the returned value is an integer but your answer ...
Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1.You may assume the array‘s length is at most 10,000.Example:Input:
[1,2,3]Output:
2Explanation:
Only two moves are needed (remember each move increments or decrements one element):[1,2,3] => [2,2...
CF1401-C. Mere Array
题意：
给出一个长度为nnn的数组aaa，你可以对这个数组进行如下操作：对于数组aaa中任意的两个元素aia_iai?、aja_jaj?，若gcd(ai,aj)=min{a1,a2,...,an}gcd(a_i,a_j)=min\{a_1,a_2,...,a_n\}gcd(ai?,aj?)=min{a1?,a2?,...,an?}，那么就可以交换数组中的这两个数字。
现在问你是否能够通过一定次的上述操作使得数组aaa变成非递减序列。思路：
首先，由于min{a1,a2,...,an}min\{a_1,a_2,...,a_n\}min{a1?,a2?,....
A zero-indexed array A of length N contains all integers from 0 to
N-1. Find and return the longest length of set S, where S[i] = {A[i],
A[A[i]], A[A[A[i]]], ... } subjected to the rule below.Suppose the first element in S starts with the selection of element
A[i] of index = i, the next element in S should be A[A[i]], and then
A[A[A[i]]]… By that analogy, we stop adding right before a duplica...
# 导包import numpy as npnumpy.array 的合并.concatenate() 一维数组x = np.array([1, 2, 3]) # array([1, 2, 3])
y = np.array([3, 2, 1]) # array([3, 2, 1])np.concatenate([x, y]) # array([1, 2, 3, 3, 2, 1])
z = np.array([666, 666, 666]) # array([666, 666, 666])np.concatenate([x, y, z])
"""
array([ 1, 2, 3, 3, 2, 1, 666, 666, 666])
""" 二维数组.concatenate((a,b,c,...),axis=0) ：默...
<?php
$arr = [‘111‘, ‘111‘, ‘111‘, ‘111‘, ‘111‘, ‘111‘, ‘111‘, ‘111‘, ‘111‘];
print_r($arr);
print_r(array_unique($arr));
print_r($arr); //array_unique后，数组本身的值并不会变?>Array
([0] => 111[1] => 111[2] => 111[3] => 111[4] => 111[5] => 111[6] => 111[7] => 111[8] => 111
)
Array
([0] => 111
)
Array
([0] => 111[1] => 111[2] => 111[3] => 11...
