zoukankan      html  css  js  c++  java
  • Codeforce 507B

    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 certain point, then rotate the circle around that pin by any angle and finally remove the pin.

    Help Amr to achieve his goal in minimum number of steps.

    Input

    Input consists of 5 space-separated integers rxyxy' (1 ≤ r ≤ 105,  - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.

    Output

    Output a single integer — minimum number of steps required to move the center of the circle to the destination point.

    Examples
    input
    2 0 0 0 4
    output
    1
    input
    1 1 1 4 4
    output
    3
    input
    4 5 6 5 6
    output
    0
    Note

    In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).

     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <vector>
     6 #include <cstdlib>
     7 #include <iomanip>
     8 #include <cmath>
     9 #include <ctime>
    10 #include <map>
    11 #include <set>
    12 using namespace std;
    13 #define lowbit(x) (x&(-x))
    14 #define max(x,y) (x>y?x:y)
    15 #define min(x,y) (x<y?x:y)
    16 #define MAX 100000000000000000
    17 #define MOD 1000000007
    18 #define pi acos(-1.0)
    19 #define ei exp(1)
    20 #define PI 3.141592653589793238462
    21 #define INF 0x3f3f3f3f3f
    22 #define mem(a) (memset(a,0,sizeof(a)))
    23 typedef long long ll;
    24 ll gcd(ll a,ll b){
    25     return b?gcd(b,a%b):a;
    26 }
    27 bool cmp(int x,int y)
    28 {
    29     return x>y;
    30 }
    31 const int N=10005;
    32 const int mod=1e9+7;
    33 int a[256];
    34 int main()
    35 {
    36     std::ios::sync_with_stdio(false);
    37     double r,x,y,x1,y1;
    38     while(scanf("%lf%lf%lf%lf%lf",&r,&x,&y,&x1,&y1)!=EOF){
    39         double dis=sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));
    40         double d=2*r;
    41         int cnt=dis/d;
    42         if(cnt*d<dis)
    43             cnt++;
    44         printf("%d
    ",cnt);
    45     }
    46     return 0;
    47 }
  • 相关阅读:
    Linux 常用指令【持续更新】
    我的 MyBatis 实现的 Dao 层
    Mybatis 点点滴滴
    Redis 初识
    Mybatis 映射关系
    Mybatis 类属性和字段映射小小分析
    Mybatis 中 sql 语句的占位符 #{} 和 ${}
    使用 Maven 管理项目
    Maven 项目依赖 pom 文件模板
    Maven 私服安装和启动
  • 原文地址:https://www.cnblogs.com/wydxry/p/7269316.html
Copyright © 2011-2022 走看看