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  • LWC 54:699. Falling Squares

    Falling Squares传送门:699.Falling SquaresProblem: On an infinite number line (x-axis), we drop given squares in the order theyThe squares are infinitely sticky on their bottom edge, and will remain fixed to any positive lengthSquares dropped adjacent to each other will not stick together prematurely.Thus, we return an answer of .Example 2: Input: , ] Output: Explanation: Adjacent squares don’t get
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  • 【LeetCode】977. Squares of a Sorted Array数组 双指针

    Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number
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  • 车道线检测--End-to-end Lane Detection through Differentiable Least-Squares Fitting

    End-to-end Lane Detection through Differentiable Least-Squares Fitting https:github.comwvangansbekeLaneDetection_End2End结合 weight maps 可以看做是 车道线的位置 x 坐标和 y 坐标,经过 least-squares layer 处理得到 车道线拟合参数2.2 Weighted Least-SquaresWeighted least-squares fitting ?2.3 Geometric Loss Function??3 Experiments3.1 Toy Experiment直线拟合 ?
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  • POJ 刷题系列:2002. Squares

    Squares传送门:2002. Squares题意: 在二维坐标系下,给出若干点,求这些点最多能组成多少个正方形。
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  • 数学知识--Methods for Non-Linear Least Squares Problems(第一章)

    Methods for Non-Linear Least Squares Problems 非线性最小二乘问题的方法 2nd Edition, April 2004 K. Madsen, H.B.Tingleff1 Introduction and Definitions 第一章 介绍和定义这本小册子主要关注 Least Squares Problem ?
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  • 数学知识--Methods for Non-Linear Least Squares Problems(第三章)

    https:blog.csdn.netzhangjunhitarticledetails88949762 Methods for Non-Linear Least Squares Problems 非线性最小二乘问题的方法Tingleff3 Non-linear least squares problems 非线性最小二乘问题 接下来我们主要关注非线性最小二乘问题的讨论。
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  • 25 Squares of a Sorted Array

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  • 做题总结—— Latin Squares

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  • 【USACO 1.2】Palindromic Squares

    进制转换,然后判断是否是回文******************************************* TASK: palsquare LANG: C++ Created Time: 2016年09月07日 星期三 21时18分46秒 ********************************* #include#include#include#includeusing namespace std; void bas(int x,int b,char ans=t%10+A; else ans=t+0; } for(int j=0;jb; for(int i=1;i
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  • 【USACO 3.2】Magic Squares

    题意4*2个格子分别为 1234 8765 的魔板有3种操作,A:上下两排互换,B:最后一列放到第一列前面,C:中间四个顺时针旋转1格。 现在给出目标状态,找出最少步数可从原始状态到达目标状态,且输出最小字典序的操作序列。题解 image.png 代码*USER:19flipp1TASK:msquareLANG:C++*#include#include#include#include#define ll long long#define in(s) freopen(#s.in,r,stdin);freopen(#s.out,w,stdout);#define N 10using namespace std;int a,fac;struct node{ int a; int d; char s;}nd;bool vis;int cantor(int a){ int ans=0,c; for(int i=1;i
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  • 数学知识--Methods for Non-Linear Least Squares Problems(第二章)

    https:blog.csdn.netzhangjunhitarticledetails88884059 Methods for Non-Linear Least Squares Problems 非线性最小二乘问题的方法
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  • LeetCode 0279 - Perfect Squares

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  • 回文平方数 Palindromic Squares

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  • Javascript错误:历史总是落后一步

    function didSomeoneWin(){ const squares = history.squares; console.log(history: , history, squares:, squares); const lines = , , , , , , , , ]; lines.forEach((line) => { const = line; if (squares &&squares === squares && squares === squares){ console.log(SOMEONE WON!); setWinner( squares); return the winner: X or O } }) } function squareClick(i){ if (!winner){ const current = history; const squares = current.squares.slice(); if (squares){ return; } squares
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  • 在返回状态访问时,React组件未更新?

    = this.state.squares.slice(); squares = X; this.setState({squares: squares}); } ); }} 总而言之,Board渲染了9个 Squares并将回调函数传递给Square,更新了Board的state。)} {this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares)}{this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares)} {this.renderSquare(this.state.squares
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  • Leetcode 840. Magic Squares In Grid

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  • Leetcode 977. Squares of a Sorted Array

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  • Leetcode 279. Perfect Squares

    Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. 计算一个数最少由多少个完全平方数相加而成。 简单DP,从1开始向上转移。class Solution {public: int numSquares(int n) { vector dp(n + 1, INT_MAX); dp = 0; for(int i = 1; i
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  • Leetcode 279. Perfect Squares

    Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. 计算一个数最少由多少个完全平方数相加而成。 简单DP,从1开始向上转移。class Solution {public: int numSquares(int n) { vector dp(n + 1, INT_MAX); dp = 0; for(int i = 1; i
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  • PKU POJ 1720 SQUARES 解题报告

    题目链接:http:acm.pku.edu.cnJudgeOnlineproblem?id=1720这题纯计算几何就搞定了,开始我写了个很长很长的代码,但是Wa掉,也不知道是代码那里有疏漏还是精度问题后来看了一篇解题报告,改为全用long处理,然后根据他的一些方法,代码量也精简了不少我的原先的方法是判断左边和底边的覆盖,分别计算每个square的覆盖区域和长度解题报告里也是计算覆盖,不过用斜率代替了我分别计算的左边和底边,同时这里我原先用的是double,但是以斜率计算可以优化为记录斜率的点,这样就不需要用double,解决了精度问题然后是判断square的重叠问题,我原先是分成两部分的,左边和底边的在后的被在前的覆盖,报告里这里有更好的算法,记录边界点,以x轴和y轴坐标和来判断哪个被哪个覆盖最后只要对每一个square,从上到下找出被覆盖的斜率,然后最后被覆盖的位置相对于圆点的斜率大于square右下角点的相对于圆点的斜率,那么这个square必然没被覆盖完全,也就是看得到我的代码:** * URL:http:acm.pku.edu.cnJudgeOnlineproblem?id=1720 * Author: OWenT * 计算几何 *#include #include #include #include using namespace std; typedef struct{ long x,y; long L;}point; point zero = {0, 0};point p; int cross( point p0, point p1 , point p2){ int t = ( p1.x -p0.x ) * ( p2.y -p0.y ) -( p2.x -p0.x ) * ( p1.y -p0.y); if ( t > 0 ) return 1; if ( t < 0 ) return -1 ; return 0 ;}bool cmp(point a , point b ){ return cross( zero , a, b ) < 0;按斜率降序排列}int main(){ long n , i , j , count; point tmp1 , tmp2; scanf(%ld, &n) ; for ( i = 0 ; i < n; i ++) { scanf(%ld %ld %ld , &p.x, &p.y , &p.L); p.y += p.L; } sort(p, p + n, cmp); count = 0 ; for ( i = 0 ; i < n; i ++) { tmp1 = p ; for ( j = 0 ; j < n; j ++) { if ( p.x + p.y 0) break;由于按斜率降序排序,第j的点斜率都小于tmp1,那么后面的必然小于 } tmp2.x = p.x + p.L ; tmp2.y = p.y - p.L ; if ( cross(zero, tmp2, tmp1 )> 0 ) count ++; } printf(%ldn, count) ; return 0;}
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