Common Tangents FZU - 2213

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：题目说是一般常见的两个圆形有四条相连接的切线
给你他们两个园的坐标及其半径，求出切线的条数
分析：分情况讨论两种圆的位置关系
重合，相交，内切，外切，内离，不相交，
依次输出
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <string.h>
#include <math.h>
#include <cmath>
using namespace std;
int main()
{
int n;
cin>>n;
int a[50];
int x1=0,y1=0,r1=0,x2=0,y2=0,r2=0;
memset(a,0,sizeof(a));
for(int i=0;i<n;i++)
{
cin>>x1>>y1>>r1>>x2>>y2>>r2;
double d=sqrt(1.0*((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
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*(r1+r2)&&d>1.0*fabs(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);
}
}