2016寒假个人赛(1)B（数学）

B - B

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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：

给出两个圆要你求共同切线，就是一道数学题目，细节较多。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<cmath>#include<ctime>#include<cstdlib>#include<queue>#include<vector>#include<set>using namespace std;const int T=1500;#define inf 0x3f3f3f3fL#define mod 1000000000typedef long long ll;typedef unsigned long long LL;int main(){#ifdef zsc    freopen("input.txt","r",stdin);#endifint N,i,j,k,u,v,x1,x2,y1,y2,r1,r2;scanf("%d",&N);while(N--){scanf("%d%d%d%d%d%d",&x1,&y1,&r1,&x2,&y2,&r2);if(x1==x2&&y1==y2){if(r1==r2){printf("-1\n");continue;}else {printf("0\n");}}double tmp = sqrt(1.0*(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));if(tmp==(r1+r2)){printf("3\n");continue;}if(tmp>(r1+r2)){printf("4\n");continue;}if(r2==(tmp+r1)||r1==(tmp+r2)){printf("1\n");continue;}if(tmp+r1<r2||tmp+r2<r1){printf("0\n");continue;}if(tmp<r1+r2){printf("2\n");continue;}}    return 0;}