Given n points on the XY plane, count how many regular rectangles are formed. A rectangle is regular if and only if its sides are all parallel to the axis.InputThe first line contains the number of tests t (1 ≤ t ≤ 10). Each case contains a single lin…

10574 - Counting Rectangles 题目链接 题意:给定一些点,求可以成几个边平行于坐标轴的矩形 思路:先把点按x排序,再按y排序.然后用O(n^2)的方法找出每条垂直x轴的边,保存这些边两点的y坐标y1, y2.之后把这些边按y1排序,再按y2排序.用O(n)的方法找出有几个连续的y1, y2都相等.那么这些边两两是能构成矩形的.为C2cnt种.然后累加起来就是答案 代码: #include <stdio.h> #include <string.h> #inc…

Description Problem H Counting Rectangles Input: Standard Input Output:Standard Output Time Limit: 3Seconds   Given n points on the XY plane, count how many regular rectanglesare formed. A rectangle is regular if and only if its sides are all paralle…

Counting Rectangles Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 1043 Accepted: 546 Description We are given a figure consisting of only horizontal and vertical line segments. Our goal is to count the number of all different rectangles…

//counting sort 计数排序 //参考算法导论8.2节 #include<cstdio> #include<cstring> #include<algorithm> #include<cassert> using namespace std; ; ; , , , , , , }; int b[n]; ]; int main() { ; memset(c, , sizeof c); ;i<n;i++) { c[a[i]]++; maxn=ma…

Counting rectangles By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles: Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the…

D. Counting Rectangles is Fun time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output There is an n × m rectangular grid, each cell of the grid contains a single integer: zero or one. Let's call t…

$ >Codeforces \space 372 B.  Counting Rectangles is Fun<$ 题目大意 : 给出一个 \(n \times m\) 的 \(01\) 矩阵,有 \(q\) 次询问,每次给出一个矩形 $ x_1, x_2, y_1, y_2$ ,求有多这个矩形的有多少个全 \(0\) 子矩形 \(1 \leq n, m \leq 50, 1 \leq q \leq 3 \times 10^5\) 解题思路 : 设 \(g(x, y, l, r)\) 表示以…

题目链接:uva 1436 - Counting heaps 题目大意:给出一个树的形状,如今为这棵树标号,保证根节点的标号值比子节点的标号值大,问有多少种标号树. 解题思路:和村名排队的思路是一仅仅的uva11174,最后问题仅仅和树德结构有直接关系.f(root)=(s(root)−1)!(s(1)∗s(2)∗⋯∗s(n) 可是给定的取模数不是质数.所以不能用逆元做.仅仅能将分子分母分别拆分成质因子,然后对质因子进制约分.由于最后的答案一定是正整数,所以对于每一个质因子,分子分解出的因子个数…

题目链接:12075 - Counting Triangles 题意:求n * m矩形内,最多能组成几个三角形 这题和UVA 1393类似,把总情况扣去三点共线情况,那么问题转化为求三点共线的情况,对于两点,求他们的gcd - 1,得到的就是他们之间有多少个点,那么情况数就能够求了,然后还是利用容斥原理去计数,然后累加出答案 代码: #include <stdio.h> #include <string.h> #include <algorithm> using nam…