Intersection (两条直线相交）
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
1
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");
}
}转载自：https://blog.csdn.net/duan_1998/article/details/54142728
