Description There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written a…
1058 - Parallelogram Counting    PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these po…
1058 - Parallelogram Counting There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points tha…
http://www.lightoj.com/volume_showproblem.php?problem=1058 题意:给你顶点,问能够成多少个平行四边形. 思路:开始想使用长度来扫描有多少根,但是好像坐标太大似乎不可行.其实我们可以通过找所有线段的中点的重合个数来计算有几个平行四边形,这种通过别的性质来判断几何关系的思维是解几何题的基础,当作入门? /** @Date : 2016-12-02-21.49 * @Author : Lweleth (SoungEarlf@gmail.com)…
1148 - Mad Counting   PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this p…
链接: https://vjudge.net/problem/LightOJ-1058 题意: There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets o…
题意: 简单的就组合数 C(m,n): 数据多,大, 要预处理: #include<bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 1e6 + 2; const LL MOD = 1000003; LL Pow_Mod(LL a, LL n) { LL ret = 1; while(n) { if(n & 1) ret = ret * a % MOD; n >>=1; a…
Parallelogram Counting Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5749   Accepted: 1934 Description There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie o…
题目来源:Light OJ 1114 Easily Readable 题意:求一个句子有多少种组成方案 仅仅要满足每一个单词的首尾字符一样 中间顺序能够变化 思路:每一个单词除了首尾 中间的字符排序 然后插入字典树 记录每一个单词的数量 输入一个句子 每一个单词也排序之后查找 依据乘法原理 答案就是每一个单词的数量之积 #include <iostream> #include <cstring> #include <cstdio> #include <algori…
题目来源:Light OJ 1429 Assassin`s Creed (II) 题意:最少几个人走全然图 能够反复走 有向图 思路:假设是DAG图而且每一个点不能反复走 那么就是裸的最小路径覆盖 如今不是DAG 可能有环 而且每一个点可能反复走 对于有环 能够缩点 缩点之后的图是DAG图 另外点能够反复走和POJ 2594一样 先预处理连通性 #include <cstdio> #include <cstring> #include <vector> #include…