福建省赛-- Common Tangents(数学几何)

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

判断两个圆有几条切线！有一点需要注意，两个圆重合的时候输出-1，重合时有无数条切线，水题一枚

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int x1,x2,y1,y2,r1,r2;
		double s1;
		scanf("%d%d%d%d%d%d",&x1,&y1,&r1,&x2,&y2,&r2);
		if(x1==x2&&y1==y2&&r1==r2)
		printf("-1\n");
		else
		{
			s1=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
			if(r1+r2<s1)
			printf("4\n");
			else if(r1+r2==s1)
			printf("3\n");
			else if(r1+r2>s1)
			{
				if(s1+min(r1,r2)==max(r1,r2))
				printf("1\n");
				else if(s1+min(r1,r2)<max(r1,r2))
				printf("0\n");
				else printf("2\n");
			}
		}
	}
	return 0;
}