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#yyds干货盘点# 解决名企真题:搬圆桌
2022-07-05 12:40:00 【51CTO】
1.简述:
描述
现在有一张半径为r的圆桌,其中心位于(x,y),现在他想把圆桌的中心移到(x1,y1)。每次移动一步,都必须在圆桌边缘固定一个点然后将圆桌绕这个点旋转。问最少需要移动几步。
输入描述:
一行五个整数r,x,y,x1,y1(1≤r≤100000,-100000≤x,y,x1,y1≤100000)
输出描述:
输出一个整数,表示答案
示例1
输入:
输出:
2.代码实现:
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n =0;
double distance =0;
while(sc.hasNext()){
// 使用long防止大数相乘溢出int范围
int r = sc.nextInt();
long x = sc.nextInt();
long y = sc.nextInt();
long x1 = sc.nextInt();
long y1 = sc.nextInt();
double a=(x1-x)*(x1-x)+(y1-y)*(y1-y);
distance=Math.sqrt(a);
n = (int)distance/(2*r);
// 若最后不能整除,需要再旋转一次
if((n*2*r)<distance){
n++;
}
System.out.println(n);
}
}
}
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