View Code of Problem 67

#include "stdio.h"
void main()
{	
	int x1,y1,x2,y2,x3,y3,x4,y4;
	double k1,k2;
	for(;scanf("%d%d%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF;)
	{
		if(x1!=x2&&x3!=x4)
		{
			k1=(y1-y2)/(x1-x2)*1.00;
			k2=(y3-y4)/(x3-x4)*1.00;
			if(k1==k2||k1==-k2)
				printf("No\n");
			else
				printf("Yes\n");
		}
		else if((x1==x2&&x3==x4))			//分母0
		{
			if(x1==x3)
				printf("Yes\n");
			else
				printf("No\n");
		}
		else
			printf("Yes\n");
		//printf("%.2lf %.2lf",k1,k2);
		

	}
}

Double click to view unformatted code.


Back to problem 67