解题思路:一开始自己想的是找出每一次旋转所得到的圆心轨迹,将想要旋转到的点代入该圆心轨迹的方程,如果相等,则跳出循环,如果不相等,则接着进行下一次旋转.后来看了题解,发现,它的旋转可以是任意角度的,所以旋转后的圆心应该在一个区域,而没有固定的圆心轨迹的方程. 如图所示,设c0为最初给出的圆,令圆c0绕O1点旋转,则对于绕O1点这一点的旋转来说,c1为其旋转后所得到的圆心轨迹,则可以看到,通过1次旋转后,距离最近为0,距离最远为线段OO2=2r, 对于第二次旋转,选取O2为圆心,作圆,再令圆c2绕…

B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem. Amr has a circle of radius r and center in point (x, …

B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem. Amr has a circle of radius r and center in point (x, …

A. Amr and Music time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an…

题目传送门 /* 贪心水题 */ #include <cstdio> #include <algorithm> #include <iostream> #include <cmath> #include <cstring> #include <vector> #include <set> #include <map> #include <string> using namespace std; ;…

Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem. Amr has a circle of radius r and center in point (x, y).…

1.CF #376 (Div. 2)    C. Socks       dfs 2.题意:给袜子上色,使n天左右脚袜子都同样颜色. 3.总结:一开始用链表存图,一直TLE test 6 (1)如果需要高效的随即存取,而不在乎插入和删除的效率,使用vector . (2)如果需要大量的插入和删除,而不关心随即存取,则应使用list . #include<bits/stdc++.h> #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a…

1.CF #375 (Div. 2)  D. Lakes in Berland 2.总结:麻烦的bfs,但其实很水.. 3.题意:n*m的陆地与水泽,水泽在边界表示连通海洋.最后要剩k个湖,总要填掉多少个湖,然后输出. #include<bits/stdc++.h> #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a,b) for (int i=a;i<=b;i++) #define mes(a,b) memset(a,b,s…

1.CF #374 (Div. 2)   D. Maxim and Array 2.总结:按绝对值最小贪心下去即可 3.题意:对n个数进行+x或-x的k次操作,要使操作之后的n个数乘积最小. (1)优先队列 #include<bits/stdc++.h> #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a,b) for (int i=a;i<=b;i++) #define mes(a,b) memset(a,b,sizeof(…

1.CF #374 (Div. 2)    C.  Journey 2.总结:好题,这一道题,WA,MLE,TLE,RE,各种姿势都来了一遍.. 3.题意:有向无环图,找出第1个点到第n个点的一条路径,经过的点数要最多. #include<bits/stdc++.h> #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a,b) for (int i=a;i<=b;i++) #define mes(a,b) memset(a,b,…