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…

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…

之前几乎没写过什么这种几何的计算题.在众多大佬的博客下终于记起来了当时的公式.嘚赶快补计算几何和概率论的坑了... 这题的要求,在对两圆相交的板子略做修改后,很容易实现.这里直接给出代码.重点的部分有:两圆在相离(或外交)时输出第一个圆的面积.内涵(或内切)则需要分类讨论,是羊的圈大.还是狼的圈大.以下是代码: #include<iostream> #include<cmath> #include<stdio.h> using namespace std; int ma…

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<…

Spell checker Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 20188   Accepted: 7404 Description You, as a member of a development team for a new spell checking program, are to write a module that will check the correctness of given word…

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]…

题解:状态压缩之后,暴力dfs,如果有一个选择,能让对手必败,那么就是必胜态,能转移到的状态都是对手的必胜态,或者无法转移,就是必败态. 总算是过了,TLE是因为状态没判重. #include<cstdio> #include<cmath> #include<vector> #include<map> #include<set> #include<algorithm> #include<iostream> #include…