2261: Triangle （皮克定理）

文章目錄

• Input
• Output
• Sample Input
• Sample Output

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	277	51	Standard

A lattice point is an ordered pair (x, y) where x and y are both integers. Given the coordinates of the vertices of a triangle (which happen to be lattice points), you are to count the number of lattice points which lie completely inside of the triangle (points on the edges or vertices of the triangle do not count).

Input

    The input test file will contain multiple test cases. Each input test case consists of six integers x1, y1, x2, y2, x3, and y3, where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of vertices of the triangle. All triangles in the input will be non-degenerate (will have positive area), and -15000 ≤ x1, y1, x2, y2, x3, y3 ≤ 15000.

    The end-of-file is marked by a test case with x1 = y1 = x2 = y2 = x3 = y3 = 0 and should not be processed.

Output

    For each input case, the program should print the number of internal lattice points on a single line.

Sample Input

0 0 1 0 0 10 0 5 0 0 50 0 0 0 0 0

Sample Output

06/*

匹克定理（s=i+b/2-1）和輾轉相除的應用

*/

#include <cstdio>
#include <iostream>
#include <memory>
#include <cmath>
using namespace std;
struct point
{
    int x,y;
}p[5];
int area()
{
    int ans=0;
    for(int i=0 ; i<3 ; i++)
    {
        ans+=p[i].x*(p[(i+1)%3].y-p[(i+2)%3].y);
        //printf("%d ",ans);
    }
    return ans;
}
int gcd(int a,int b)
{
 return b==0?a:gcd(b,a%b);
}
int atline (point p1,point p2)
{
    int b=fabs(p1.x-p2.x) , a=fabs(p1.y-p2.y);
    return gcd(a,b);//返回邊上點的個數加上一個端點
}
int main ()
{
    int i,n;
    while (1)
    {
        bool flag=1;
        for ( i=0 ; i<3 ; i++)
        scanf("%d%d",&p[i].x,&p[i].y);
        for ( i=0 ; i<3 ; i++)
         if ( p[i].x || p[i].y )flag=0;
        if(flag) break;
        int s=fabs(area())+2,side=0;
        //cout<<s<<"/n";
        for (i=0 ; i<3 ; i++)
        {
            side+=atline(p[i],p[(i+1)%3]);
        }
        int ans=(s-side)/2;
        printf("%d/n",ans);
    }
    return 0;
}
/*
0 0 1 0 0 1
0 0 5 0 0 5
5 5 5 0 0 5
0 0 0 0 0 0
*/