Codeforces Beta Round #21 / 21B Intersection(数学&详细分情况)

解析代码片段：两组点集交集数量计算

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本文详细解析了一个代码片段，用于计算由两条直线方程定义的两组点集的交集数量。包括输入参数解析、特殊情况处理、等价点集判断及最终输出结果的实现。

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.

   Sample test(s)
 
     input
   
1 1 0
2 2 0

     output
   
-1

     input
   
1 1 0
2 -2 0

     output
   
1

反思：一定要先判断特殊情况！

1. 题目说的是两个点集。

2. 优先判断特殊情况——某一集合是空集或（该题定义下的）全集。

3. 排除上述情况后，判断两个点集是否等价即可。

完整代码：

/*15ms,0KB*/

#include<cstdio>
using namespace std;

int main()
{
	int  a, b, c, A, B, C;
	scanf("%d%d%d%d%d%d", &a, &b, &c, &A, &B, &C);
	if ((a == 0 && b == 0) || (A == 0 && B == 0))
	{
		if ((a == 0 && b == 0 && c != 0) || (A == 0 && B == 0 && C != 0))
			printf("0");///空集
		else
			printf("-1");///全集
	}
	else if (a * B == A * b)
	{///看两个点集是否等价，注意当x前系数为0时改用y前系数判断
		if (a)
			printf("%d", a * C != A * c ? 0 : -1);
		else
			printf("%d", b * C != B * c ? 0 : -1);
	}
	else
		printf("1");
	return 0;
}