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RSS订阅原创 相位小数偏差（UPD）估计基本原理
PPP中的一个关键性难题在于非差模糊度固定，成功固定非差模糊度可以使 PPP。由于估计时存在一个秩亏，需要选择一颗卫星或一个测站的UPD作为基准以消除秩亏。参考论文：《基于不同PPP模型的北斗三号相位小数偏差估计与新特性分析》是致使相位模糊度失去整数特性的主要因素，精确估计并校正 UPD。是实现非差模糊度固定的重要前提，也是实现GNSS。的收敛速度和定位精度得到显著提升。相位小数偏差 (UPD)
2024-12-15 23:04:24
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原创 openmp报错：内存资源不足，无法创建线程
OMP: System error #8: 内存资源不足，无法处理此命令。原因为栈的保留提交大小过大，创建多线程时导致内存资源不足。
2024-12-04 15:32:16
176
原创 Linux服务器基于NFS实现共享目录
NFS简介：NFS（Network File System）是一种分布式文件系统协议，允许用户通过网络访问远程计算机上的文件和目录，就像访问本地文件一样。5、服务器B安装 nfs-utils、rpcbind，启用rpcbind。应用场景：服务器A将某目录设置为共享目录，服务器B可访问并写入。1、服务器A安装nfs-utils、rpcbind 软件包。6、服务器B创建目录，并挂载至服务器A的共享目录。2、服务器A设置共享目录与可访问的服务器。4、服务器A发布的NFS共享目录。
2024-08-09 16:39:17
426
原创 Segmentation fault问题解决
在Linux服务器中出现forrtl: severe（174）：SIGSEGV, segmentation fault occurred。取消堆栈限制后，程序可正常运行。
2024-03-27 17:30:21
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原创 启用NTP服务解决Linux系统时间与北京时间不同步问题
1、服务器的Linux版本为Linux version 4.18.0-348.7.1.el8_5.x86_64 (mockbuild@kbuilder.bsys.centos.org) (gcc version 8.5.0 20210514 (Red Hat 8.5.0-4) (GCC)) #1 SMP Wed Dec 22 13:25:12 UTC 2021。2、NTP即Network Time Protocol（网络时间协议），是一个互联网协议，用于同步计算机之间的系统时钟。
2023-11-03 16:32:01
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原创 GNSS卫星姿态解算
卫星的姿态指卫星的星体坐标轴XYZ在惯性系下的指向，其中Z轴始终指向地球地心，Y轴为卫星太阳能帆板的旋转轴，它始终与太阳-卫星方向垂直（为了最大程度地利用太阳辐射能量）。当然，Y轴同时也垂直于Z轴，最后的X轴与Y轴和Z轴组成右手坐标系。以上是卫星在非地影期间的姿态解算，然而在卫星地影期间则需考虑更多因素，首先需要求解太阳高度角β与卫星轨道角μ，其中β可以由轨道面法向向量W与xsun求得，在非地影期的数据解算过程中，我们已知惯性系下的太阳位置向量xsun，卫星位置向量xsat，卫星速度向量vsat。
2023-10-31 23:22:28
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原创 Leetcode43：接雨水（动态规划）
解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]输入：height = [4,2,0,3,2,5]...
2022-08-29 17:47:21
529
原创 滑动窗口模板（leetcode76 最小覆盖子串）
给你一个字符串 s 、一个字符串 t。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 ""。对于 t 中重复字符，我们寻找的子字符串中该字符数量必须不少于 t 中该字符数量。输入：s = "ADOBECODEBANC", t = "ABC"解释: t 中两个字符 'a' 均应包含在 s 的子串中，如果 s 中存在这样的子串，我们保证它是唯一的答案。输入: s = "a", t = "aa"输入：s = "a", t = "a"...
2022-08-28 21:53:35
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原创 C++remove和erase
remove(str.begin(),str.end(),val);remove是将strz中等于val的值用后一个值覆盖，并返回最后一个未被删除的元素的下一个迭代器：若str为"abcaad"对'a'remove后则变为"bcdaaa"配合erase可将所有a删除str.erase(remove(str.begin(),str.end(),val),str.end());...
2022-04-05 23:40:13
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原创 C++求解最大公约数和最小公倍数
最大公约数（greatest common diversor）int gcd(int x,int y){ if(y%x==0) return x; return gcd(y%x,x);}最小公倍数（lowest common multiple）int lcm(int x,int y){ return x*y/gcd(x,y);}
2022-04-05 19:32:22
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原创 卫星广播星历格式
1、GPS广播星历G01 2022 01 04 00 00 00 4.665437154472E-04-1.000444171950E-11 0.000000000000E+00 7.300000000000E+01-4.275000000000E+01 4.347323940635E-09-5.174246376131E-01 -2.102926373482E-06 1.122187799774E-02-4.135072231293E-07 5.153668924332E+03...
2022-03-15 15:22:27
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原创 剑指 Offer 10- II. 青蛙跳台阶问题(矩阵快速幂)
class Solution {public: typedef vector<vector<long>> vvl; vvl multiply(vvl& a,vvl& b){ vvl c={{1,0},{0,1}}; for(int i=0;i<2;i++){ for(int j=0;j<2;j++){ c[i][j]=(a[i][...
2022-02-19 21:02:13
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原创 【代码模板】快速幂（递归+迭代）+矩阵快速幂（递归+迭代）
class Solution {public: double myPowHelper(double x, long n){ if(n==0) return 1; double y=myPowHelper(x,n/2); return n&1?y*y*x:y*y; } double myPow(double x, int n) { long n_long=n;//INT_MIN绝对值比INT_MAX大1，取反.
2022-02-19 16:41:15
239
原创 剑指Offer004：只出现一次的数字
方法一：哈希表，需要额外空间class Solution {public: int singleNumber(vector<int>& nums) { unordered_map<int,int> hash_map; for(auto& num:nums){ hash_map[num]++; } for(auto& [val,count]:hash_ma..
2022-02-19 13:41:25
121
原创 剑指Offer003：前 n 个数字二进制中 1 的个数
方法一：bitset容器class Solution {public: vector<int> countBits(int n) { vector<int> ans; for(int i=0;i<=n;i++){ bitset<32> bs(i); ans.push_back((int)bs.count()); } return ans..
2022-02-19 13:16:06
374
原创 剑指Offer002：二进制加法
class Solution {public: string addBinary(string a, string b) { int idx_a=a.size()-1,idx_b=b.size()-1; string ans=""; int carry=0;//进位初始化为0 while(idx_a>=0||idx_b>=0||carry!=0){ int sum=carry; ...
2022-02-19 12:15:39
7536
原创 剑指Offer001:整数除法
提示:-231<= a, b <= 231- 1 b != 0class Solution {public: int divide(int a, int b) { bool sign=(a^b)>>31;//判断a和b是否异号，右移31位到符号位 long dividend=abs((long)a),divisor=abs((long)b);//转为正数 long ans=0; while(d...
2022-02-19 12:11:31
312
原创 Leetcode688:骑士在棋盘上的概率（动态规划）
On an n x n chessboard, a knight starts at the cell (row, column) and attempts to make exactly k moves. The rows and columns are 0-indexed, so the top-left cell is (0, 0), and the bottom-right cell is (n - 1, n - 1).A chess knight has eight possible move
2022-02-17 13:19:28
482
原创 Leetcode1719:重构一棵树的方案数（拓扑）
You are given an array pairs, where pairs[i] = [xi, yi], and:There are no duplicates.xi < yiLet ways be the number of rooted trees that satisfy the following conditions:The tree consists of nodes whose values appeared in pairs.A pair [xi, yi] exi
2022-02-16 15:08:11
327
原创 Leetcode297:二叉树的序列化与反序列化（广搜）
Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer env
2022-02-15 20:14:12
491
原创 Leetcode236:二叉树的最近公共祖先（广搜）
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as d
2022-02-14 21:33:08
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原创 Leetcode116:填充每个节点的下一个右侧节点指针（层序遍历）
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:struct Node { int val; Node *left; Node *right; Node *next;}Populate each next pointer t...
2022-02-12 16:58:57
225
原创 Leetcode105:从前序与中序遍历序列构造二叉树（哈希表+递归）
Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.Constraints:1 <= preorder.length <= 3000inorder.l
2022-02-12 15:55:29
209
原创 Leetcode106:从中序与后序遍历序列构造二叉树（哈希表+递归）
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.Constraints:1 <= inorder.length <= 3000posto
2022-02-12 15:37:16
342
原创 Leetcode1020:飞地的数量（广搜）
You are given an m x n binary matrix grid, where 0 represents a sea cell and 1 represents a land cell.A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the grid.Return the numbe
2022-02-12 14:44:57
327
原创 Leetcode1984:学生分数的最小差值（排序+划窗）
You are given a 0-indexed integer array nums, where nums[i] represents the score of the ith student. You are also given an integer k.Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k score
2022-02-11 11:22:10
519
原创 Leetcode2006:差的绝对值为K的数对数目（哈希表）
Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.The value of |x| is defined as:x if x >= 0.-x if x < 0.Constraints:1 <= nums.length <= 200 1 <= num
2022-02-09 12:11:39
5979
原创 Leetcode1405:最长快乐字符串（贪心）
A string s is called happy if it satisfies the following conditions:s only contains the letters 'a', 'b', and 'c'.s does not contain any of "aaa", "bbb", or "ccc" as a substring.s contains at most a occurrences of the letter 'a'.s contains at most b o
2022-02-07 17:39:23
610
原创 Leetcode1748:唯一元素的和（哈希表）
You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.Return the sum of all the unique elements of nums.Constraints:1 <= nums.length <= 100 1 <= nums[i] <= 100clas
2022-02-06 17:30:11
339
原创 Leetcode1219:黄金矿工（DFS+lambda表达式）
In a gold mine grid of size m x n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.Return the maximum amount of gold you can collect under the conditions:Every time you are located in a cell you will
2022-02-05 12:11:32
1543
原创 Leetcode1414:和为K的最小斐波拉契数字数目（贪心+二分）
Given an integer k, return the minimum number of Fibonacci numbers whose sum is equal to k. The same Fibonacci number can be used multiple times.The Fibonacci numbers are defined as:F1 = 1F2 = 1Fn = Fn-1 + Fn-2 for n > 2.It is guaranteed that for
2022-02-03 16:07:07
766
原创 Leetcode1349:参加考试的最大学生数（动态规划+状态压缩+位运算）
Given a m * n matrix seats that represent seats distributions in a classroom. If a seat is broken, it is denoted by '#' character otherwise it is denoted by a '.' character.Students can see the answers of those sitting next to the left, right, upper left
2022-02-02 17:22:18
615
毕业论文大礼包.rar
2020-07-18
白塞尔大地主体解算（正算+反算）
2018-06-15
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