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). He wants the circle center to be in new position (x', y'). In one step Amr can put a pin to the border of the circle in a…

题目链接:http://codeforces.com/problemset/problem/507/B 题目意思:给出圆的半径,以及圆心坐标和最终圆心要到达的坐标位置.问最少步数是多少.移动见下图.(通过圆上的边固定转动点,然后转动任意位置,圆心就会移动了,这个还是直接看图吧) 解题的思路就是,两点之间,距离最短啦----要想得到最少步数,我们需要保证圆心在这条连线上移动,每次转动的角度是180度,而每步移动的距离是2r,直到两个圆交叉,要注意最后一步转动的角度可能会小于180度.最后就是注意精…

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, …

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). He wants the circle center to be in new position (x', y'). In one step Amr can put a pin to the border of the circle in a…

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, …

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).…

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…

C. Amr and Chemistry Problem's Link: http://codeforces.com/problemset/problem/558/C Mean: 给出n个数,让你通过下面两种操作,把它们转换为同一个数.求最少的操作数. 1.ai = ai*2 2.ai = ai/2 (向下取整) analyse: 基本思路:首先枚举出每个数能够到达的数字并且记录下到达该数组需要的步数,然后从到达次数为n次的数字中选择步数最小的即为答案. 对于一个数字Ai,它可以变换得到的数字可…

题目链接:http://codeforces.com/problemset/problem/558/B 题目意思:给出一个序列,然后找出出现次数最多,但区间占用长度最短的区间左右值. 由于是边读入边比较,因此问题最关键的是,记录每个数第一次出现的位置,即左值.因为要保证次数是出现最多,因此需要一个cnt[]数组来记录出现次数.然后当最多出现次数与当前cnt[x]次数相同时,要选择区间较短的,再更新左右区间值. 赛中短路竟然想不出来~~~泪啊~~泪啊- >_< #include <iost…