POJ2954 Triangle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5869		Accepted: 2524

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).

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

Sample Output

0
6

Source

Stanford Local 2004

给定三个点的坐标，问三个点连成的三角形中，包含了多少整坐标点

数学问题计算几何？ pick定理

pick定理：设平面直角坐标系中，一个三角形内部有a个整坐标点，它的三条边共经过了b个整坐标点，那么三角形面积S=a + b/2 +1

把公式变形一下，用海伦公式或者叉积求出S，就能算出a了

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 int read(){
 9     int x=0,f=1;char ch=getchar();
10     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 struct point{
15     int x,y;
16     point operator - (point rhs)const{
17         return (point){x-rhs.x,y-rhs.y};
18     }
19     int cross(point rhs){return x*rhs.y-y*rhs.x;}
20 }p[4];
21 inline int abs(int x){return x<0?-x:x;}
22 int gcd(int a,int b){return (!b)?a:gcd(b,a%b);}
23 int area(point a,point b,point c){
24     return abs((c-a).cross(b-a));
25 }
26 int side(){
27     p[0]=p[3];int res=0;
28     for(int i=1;i<=3;i++){
29         res+=gcd(abs(p[i].x-p[i-1].x),abs(p[i].y-p[i-1].y));
30     }
31     return res;
32 }
33 int main(){
34     int i,j;
35     while(scanf("%d%d%d%d%d%d",&p[1].x,&p[1].y,&p[2].x,&p[2].y,&p[3].x,&p[3].y)!=EOF){
36         if(!p[1].x && !p[2].x && !p[3].x && !p[1].y && !p[2].y && !p[3].y)break;
37         int ans=area(p[1],p[2],p[3])+2-side();
38         printf("%d
",ans/2);
39     }
40     return 0;
41 }