第六届福建省大学生程序设计竞赛-重现赛，Problem B Common Tangents【数学几何】

最新推荐文章于 2020-03-08 18:52:07 发布

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最新推荐文章于 2020-03-08 18:52:07 发布

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本文链接：https://blog.youkuaiyun.com/liu6886/article/details/50419570

本文介绍了一道关于计算两个圆共有切线条数的问题，通过判断两圆位置关系（如相离、相交等），给出了AC代码实现，并详细解释了每种情况下公切线的数量。

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Problem B Common Tangents

Accept: 191 Submit: 608
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Two different circles can have at most four common tangents.

The picture below is an illustration of two circles with four common tangents.

Now given the center and radius of two circles, your job is to find how many common tangents between them.

Input

The first line contains an integer T, meaning the number of the cases (1 <= T <= 50.).

For each test case, there is one line contains six integers x1 (−100 ≤ x1 ≤ 100), y1 (−100 ≤ y1 ≤ 100), r1 (0 < r1 ≤ 200), x2 (−100 ≤ x2 ≤ 100), y2 (−100 ≤ y2 ≤ 100), r2 (0 < r2 ≤ 200). Here (x1, y1) and (x2, y2) are the coordinates of the center of the first circle and second circle respectively, r1 is the radius of the first circle and r2 is the radius of the second circle.

Output

For each test case, output the corresponding answer in one line.

If there is infinite number of tangents between the two circles then output -1.

Sample Input

3

10 10 5 20 20 5

10 10 10 20 20 10

10 10 5 20 10 5

Sample Output

4

2

3

题意：求两个圆的切线有几条；

思路，求出两个圆心之间的距离，比较5种情况：重合输出-1 无数条，相离 4 ，外切 3 ，相交 2，内切 1，内含 0.

ac-code：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		int x1, y1, r1, x2, y2, r2;
		scanf("%d%d%d%d%d%d", &x1, &y1, &r1, &x2, &y2, &r2);
		double d;
		d = sqrt( 1.0*((x1-x2)*(x1-x2) + (y2-y1)*(y2-y1)) );//sqrt会损失精度，以后可以把要比较的平方 
		int a;
		if(d==0.0 && r1==r2)	a = -1;
		else  if(d > 1.0*(r1+r2))	a = 4;
		else  if(d==1.0*(r1+r2))	a = 3;
		else  if(d>1.0*(fabs(r1-r2)) && d < 1.0*(r1+r2))	a = 2;
		else  if(d==1.0*(fabs(r1-r2)))  a = 1;
		else  if(d < 1.0*(fabs(r1-r2))) a = 0;
		printf("%d\n", a);
	}
	return 0;	
}