传送门 C. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has…

A. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a re…

Codeforces Round #296 (Div. 1)C. Data Center Drama Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/528/problem/C Description The project of a data center of a Big Software Company consists of n compu…

A. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a re…

A. Playing with Paper 如果a是b的整数倍,那么将得到a/b个正方形,否则的话还会另外得到一个(b, a%b)的长方形. 时间复杂度和欧几里得算法一样. #include <iostream> #include <cstdio> using namespace std; //const int maxn = ; int main() { //freopen("in.txt", "r", stdin); ; scanf(&q…

http://codeforces.com/contest/528/problem/E 先来吐槽一下,一直没机会进div 1, 马力不如当年, 这场题目都不是非常难,div 2 四道题都是水题! 题目大意:给n条直线,保证直线两两不平行,保证三条直线不公点.然后,随机挑三条直线,构成一个三角形,问挑出的三角形的面积的期望. 换句话说,就是算出全部三角形的面积和.再除以三角形的数量.后者是C(n,3), 关键是三角形面积和怎样计算. 下面给我我的思路:O(n^2) 基于三角形的向量表示法, S(A…

题目传送门 /* 无算法 三种可能:1.交换一对后正好都相同,此时-2 2.上面的情况不可能,交换一对后只有一个相同,此时-1 3.以上都不符合,则不交换,-1 -1 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map>…

题目传送门 /* 水题 a或b成倍的减 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <vector> #include <set&…

A. Playing with Paper One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm  ×  b mm sheet of paper (a > b). Usually the first step in making an origami is making a square piece of paper from the…

A:模拟辗转相除法时记录答案 B:3种情况:能降低2,能降低1.不能降低分别考虑清楚 C:利用一个set和一个multiset,把行列分开考虑.利用set自带的排序和查询.每次把对应的块拿出来分成两块插入回去.然后行列分别取最大相乘的作为这次询问的答案 D:一个区间覆盖问题的变形.注意公式的话.非常easy发现事实上x.w相应的就是一个[x - w, x + w]的区间,然后求最多不重合区间就可以 代码: #include <cstdio> #include <cstring> #…