World CupInput file: Standard InputOutput file: Standard OuptutTime limit: 1 second Here is World Cup again, the top 32 teams come together to fight for the World Champion.The teams are assigned into 8 groups, with 4 teams in each group. Every two te…

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5930 题意: 现有四支队伍两两打比赛,总共就是打六场比赛,每场比赛赢的队伍可得 $3$ 分,输的队伍得 $0$ 分,平局则两个队各得 $1$ 分. 现在给出四个队伍最终的积分…

Hidden Code 题目连接: http://codeforces.com/gym/100015/attachments Description It's time to put your hacking skills to the test! You've been called upon to help crack enemy codes in the current war on... something or another. Anyway, the point is that yo…

A. ArielTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/A Description King Triton really likes watching sport competitions on TV. But much more Triton likes watching live competitions. So Triton decides to set up…

G. #TheDress Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/G Description After landing on planet i1c5l people noticed that blue and black clothes are quite popular among the locals. Each aboriginal has at least…

http://codeforces.com/gym/101055/problem/A 题目:给定一些三维空间的点,要你找一个平面,能覆盖尽量多的点,只要求输出点数即可.n<=50 因为数据量小,我们考虑暴力. 首先,三个不在同一条直线的点,确定一个平面,然后枚举其他的点.判断一下,这样的复杂度是n^4.可以接受 特判.所有点都在同一条直线.直接输出n. 后面的,枚举三个点后,能算出这个平面的法向量,然后枚举其他点,与法向量的数量积是0的,就可以ans++ #include <cstdio>…

原题链接:http://codeforces.com/gym/100203/attachments/download/1702/statements.pdf 题解 考虑暴力的复杂度是O(n^3),所以我们需要记录所有的ai+aj,如果当前考虑到了ak,那么就去前面寻找ai,使得ak-ai是我们记录过的和.整个算法的复杂度O(n^2). 代码 #include<iostream> #include<cstring> #include<cstdio> #include<…

Identity Checker 题目连接: http://codeforces.com/gym/100015/attachments Description You likely have seen that x(sin x +cos2 x) ! x = 0, and you may have seen that sin(2x) ! 2 sin x cos x =0. But did you know that tan (2x)(x ! x tan2 x) ! 2x tan x = 0? Wo…

题目链接:http://codeforces.com/gym/101848/problem/B 给出一串数字要你最多改动三个数字使这一串数字成为等差数列.因为最多改动三个数字所以可以先求出相邻两项的差值再进行修改,判断是否符合,最多循环四遍即可. #include<iostream> using namespace std; int main() { ],b[]; cin>>n; ;i<n;i++) cin>>a[i]; ) { ;i<n;i++) b[i]…

因为只能买一次,暴力枚举一下买的衣服的大小. #include<cstdio> #include<map> #include<algorithm> using namespace std; typedef long long ll; #define fi first #define se second ; map<int,int> S; int main() { int T; scanf("%d",&T); ; k <= T…