#pragmaGCCoptimize("O3,unroll-loops")//#pragmaGCCtarget("avx2,bmi,bmi2,lzcnt,popcnt")//如果在不支持avx2的平台上将avx2换成avx或SSE之一#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int

区间众数要求即有次数又要数字最小#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedef

一开始默认为0，如果有捕食关系调整d[x]#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;ty

方法1：带权路径维护本题核心：[a,b]之间有奇数个1转换为s[a-1]^s[b]=1，从而转向并查集#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedint

考虑利润从高到底排序，优先把利润最高的填到最接近过期的空余位置贪心证明：如果当前a[i]被填入而填入了比最靠近过期前面的位置显然不会更优，如果没有被填入而放了另一组方案利润更大，则把属于a[i]的位置换成a[i]一定更优，两者矛盾#include<bits/stdc++.h>usingnamespacestd;#def

相等放并查集里，不等直接判断#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefv

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

#include<bits/stdc++.h>usingnamespacestd;#definexfirst#defineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvector<string>VS;typedefvector<int>

include<bits/stdc++.h>usingnamespacestd;definexfirstdefineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvectorVS;typedefvectorVI;typedefvector<vect

include<bits/stdc++.h>usingnamespacestd;definexfirstdefineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvectorVS;typedefvectorVI;typedefvector<vect

include<bits/stdc++.h>usingnamespacestd;definexfirstdefineysecondtypedefpair<int,int>PII;typedeflonglongll;typedefunsignedlonglongull;typedefunsignedintuint;typedefvectorVS;typedefvectorVI;typedefvector<vect

在OpenVX中添加自定义节点大概通过以下步骤实现：定义自定义节点的计算逻辑：你需要编写一个C函数来实现自定义的图像处理操作。创建自定义节点：通过定义一个自定义节点核（kernel），并将其注册到OpenVX上下文中。在图中使用自定义节点：使用你定义的节点与OpenVX提供的内

根据点坐标X，Y，Z和方向向量Vx，Vy，Vz绘制射线簇1.数据导入paraview可以直接导入以逗号分隔的txt或csv数据，但无法自动识别，需要使用过滤器转为paraview可以操作的数组。以下是以空格分隔的txt数据为例，paraview5.11.1版本。数据的存储格式如下图所示，以空格隔开。x，y，z为点的空间坐标，vx，v

写在前面：

引入扫描线一般运用在图形上面，它和它的字面意思十分相似，就是一条线在整个图上扫来扫去，它一般被用来解决图形面积，周长，以及二维数点等问题。Atlantis问题题意在二维坐标系上，给出多个矩形的左下以及右上坐标，求出所有矩形构成的图形的面积。解法根据图片可知总面积可以直接暴力

题目描述给定一个序列\(a_1,a_2,\dots,a_n\)，\(m\)个询问。每个询问指定一个区间\([l,r]\)，你需要输出\(a_l,a_{l+1},\dots,a_r\)这些数字里出现次数最多的数的出现次数。输入第一行一个整数\(T(1\leqT\leq6)\)，表示测试数据的组数。每组数据第一行两个数\(n,m(1\leqn,m\leq

对于直方图的比较，我们可以使用OpenCV提供的函数compareHist()进行比较，从而得到一个数值，表示两个直方图的匹配程度（相似性）。原理对于两个直方图（H1

  importturtleimportmathfromrandomimportuniformfromdataclassesimportdataclassWIDTH=1200HEIGHT=800MIN_V=5MAX_V=15MIN_SIZE_FACTOR=0.7MAX_SIZE_FACTOR=4START_DISTANCE=600R=10MARGIN=50SLEEP_MS=20@dataclassclas