# 皮克定理及其应用

## H - 三角形

4.1 Description

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).

4.2 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. For example:

0 0 1 0 0 1
0 0 5 0 0 5
0 0 0 0 0 0

4.3 Output

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

0
6

2*S=2*A+B-2
(S为三角形面积，A为三角形内部的整点数，B为三角形边上整点数）

a=(2S-b+2)/2

1 #include <iostream>
2 #include<cmath>
3 #include<cstdlib>
4 #include<cstring>
5 #include<string>
6 #include <algorithm>
7 using namespace std;
8 typedef long long ll;
9
10 struct point
11 {
12     int x,y;
13 }p[5];
14 int area()
15 {
16     int ans=0;
17     for(int i=0 ; i<3 ; i++)
18     {
19        ans+=p[i].x*(p[(i+1)%3].y-p[(i+2)%3].y);
20     }
21      return ans;
22 }
23 int gcd(int a,int b)
24 {
25     return b==0?a:gcd(b,a%b);
26 }
27 int atline (point p1,point p2)
28 {
29     int b=fabs(p1.x-p2.x) , a=fabs(p1.y-p2.y);
30     return gcd(a,b);
31 }
32 int main ()
33 {
34     bool flag;
35     int i,n,s,lpoint;
36      while (1)
37     {
38         flag=1;
39         for ( i=0 ; i<3 ; i++)
40             scanf("%d%d",&p[i].x,&p[i].y);
41         for ( i=0 ; i<3 ; i++)
42         if ( p[i].x || p[i].y )flag=0;
43         if(flag) break;
44         s=fabs(area())+2;
45         for (i=0 ; i<3 ; i++)
46         {
47             lpoint+=atline(p[i],p[(i+1)%3]);
48         }
49         int ans=(s-lpoint)/2;
50         printf("%d\n",ans);
51     }
52     return 0;
53 }
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