HDU 5839 Special Tetrahedron 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5839 Description Given n points which are in three-dimensional space(without repetition). Please find out how many distinct Special Tetrahedron among them. A tetrahedron is ca…

Special Tetrahedron 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5839 Description Given n points which are in three-dimensional space(without repetition). Please find out how many distinct Special Tetrahedron among them. A tetrahedron is called Sp…

Special Tetrahedron 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5839 Description Given n points which are in three-dimensional space(without repetition). Please find out how many distinct Special Tetrahedron among them. A tetrahedron is called Sp…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5839 在一个三维坐标,给你n个点,问你有多少个四面体(4个点,6条边) 且满足至少四边相等 其余两边不相邻. 暴力4重循环,但是在第3重循环的时候需要判断是否是等腰三角形,这便是一个剪枝.在第4重循环的时候判断4点是否共面 (叉乘), 5或者6边相等就+1,4边相等就判断另外两边是否相交就行了. 赛后过的,觉得自己还是太菜了. //#pragma comment(linker, "/STACK:10…

CCPC网络赛 HDU5839 Special Tetrahedron 题意:n个点,选四个出来组成四面体,要符合四面体至少四条边相等,若四条边相等则剩下两条边不相邻,求个数 思路:枚举四面体上一条线,再找到该线两个端点相等的点,放在一个集合里面. 要符合条件的话,则该集合里面找两个点,并且要判断一下. 注意,普通四面体会被重复计算两次,正四面体会重复计算六次 #include <bits/stdc++.h> using namespace std; #define LL long long…

Special Tetrahedron Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 328    Accepted Submission(s): 130 Problem Description Given n points which are in three-dimensional space(without repetition)…

空间的200个点,求出至少四边相等,且其余两边必须不相邻的四面体的个数. 用map记录距离点i为d的点有几个,这样来优化暴力的四重循环. 别人的做法是枚举两点的中垂面上的点,再把到中点距离相等的点找出来,n^3的样子. 还要注意四个点共面的情况. 共面判断就是用叉乘计算出ijk三点所在面的法向量,然后判断il向量是否和法向量垂直,是则共面. #include <cstdio> #include <cstring> #include <algorithm> #includ…

Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4569 Description Let f(x) = a nx n +...+ a 1x +a 0, in which a i (0 <= i <= n) are all known integers. We call f(x) 0 (mod…

Special Fish Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2367    Accepted Submission(s): 878 Problem Description There is a kind of special fish in the East Lake where is closed to campus o…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4569 题意:给你一个最高幂为4的一元多项式,让你求出一个x使其结果模p*p为0. 题解:f(x)%(p*p)=0那么一定有f(x)%p=0,f(x)%p=0那么一定有f(x+p)%p=0. 所以我们可以开始从0到p枚举x,当f(x)%p=0,然后再从x到p*p枚举,不过每次都是+p,找到了输出即可,没有的话No solution! AC代码: #include <iostream> #includ…