Intersection (两条直线相交）

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最新推荐文章于 2025-05-30 22:11:42 发布

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本文介绍了一种通过编程解决两组由线性方程确定的点集交集数量的方法。根据输入的两个线性方程参数，程序能够判断这两条直线是否相交、平行或者重合，并据此输出交点的数量或是无限多点的标识。

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You are given two set of points. The first set is determined by the equation A₁x + B₁y + C₁ = 0, and the second one is determined by the equation A₂x + B₂y + C₂ = 0.

Write the program which finds the number of points in the intersection of two given sets.

Input

The first line of the input contains three integer numbers A₁, B₁, C₁ separated by space. The second line contains three integer numbers A₂, B₂, C₂ separated by space. All the numbers are between -100 and 100, inclusive.

Output

Print the number of points in the intersection or -1 if there are infinite number of points.

Example

Input

1 1 0
2 2 0

Output

-1

Input

1 1 0
2 -2 0

Output

思路就是用乘法逃过除数为零的尴尬。讨论各种情况，直线不存在，相交。平行。重合，无穷多个点。

代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
    int a,b,c,aa,bb,cc;
    while(~scanf("%d%d%d%d%d%d",&a,&b,&c,&aa,&bb,&cc))
    {
        if(!a&&!b&&c||!aa&&!bb&&cc)
        {
            printf("0\n");
            return 0;
        }
        if(a*bb==b*aa)
        {
            if(!a&&!aa)
            {
                if(b*cc==c*bb)
                    printf("-1\n");
                else
                    printf("0\n");

            }
            else
            {
                if(a*cc==c*aa)
                    printf("-1\n");
                else
                    printf("0\n");

            }
        }
        else
            printf("1\n");
    }
}