Common Tangents【第六届福建省大学生程序设计竞赛-重现赛】

最新推荐文章于 2019-08-10 21:49:54 发布

原创最新推荐文章于 2019-08-10 21:49:54 发布 · 957 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM

ACM解题报告同时被 2 个专栏收录

40 篇文章

订阅专栏

ACM__数学

6 篇文章

订阅专栏

本篇介绍了一个算法问题，即给定两个圆的圆心坐标和半径后，如何确定它们之间的共切线数量。文章通过数学方法讨论了所有可能的情况，包括4条、3条、2条、1条、0条以及无限条的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Common Tangents

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on FZU. Original ID: 2213
64-bit integer IO format: %I64d Java class name: Main

Submit Status Statistics Discuss

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：给定两个圆心的坐标和半径

输出有多少个相切的切线

一共有 6种可能 4条 3条 2条 1条 0条无限条（输出-1）

数学问题考虑到所有情况就好了

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main (void)
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int x1,x2,y1,y2,r1,r2;
        int lang,maxr,minr,min,max;
        int x,y;
        scanf("%d%d%d%d%d%d",&x1,&y1,&r1,&x2,&y2,&r2);
        x=abs(x1-x2);
        y=abs(y1-y2);
        lang=x*x+y*y;
        double lang2=sqrt(lang);
        //printf("%lf",lang2);
        if(r1>=r2)
        {
            maxr=r1;
            minr=r2;
        }
        else
        {
            maxr=r2;
            minr=r1;
        }
        max=maxr+minr;
        min=maxr-minr;
        if(lang>max*max)
        {
            printf("4\n");
            continue;
        }
        else if(lang==max*max)
        {
            printf("3\n");
            continue;
        }
        else if(x1==x2&&r1==r2&&y1==y2)//这一步要放在1条的上面 
        {
            printf("-1\n");
            continue;
        }
        else if(lang==min*min)
        {
            printf("1\n");
            continue;
        }

        else if(lang<max*max&&lang2+minr>maxr)//相交注意有2个条件
        {
            printf("2\n");
            continue;
        }
        else if(lang2+minr<maxr)
        {
            printf("0\n");
            continue;
        }
    }
    return 0;
}