poj 2954 Triangle（Pick定理）

 

Triangle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5546		Accepted: 2410

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

 

【思路】

       Pick定理

       定理如下：

              S=a+b/2-1

       其中S代表多边形面积，a代表多边形内部的整点数，b代表多边形边上的整点数。

       其中b=gcd(|x2-x1|,|y2-y1|)

 

【代码】

 

#include<cstdio>
using namespace std;

int gcd(int a,int b) {
    return b==0? a:gcd(b,a%b);
}
int x1,y1,x2,y2,x3,y3;
int abs(int x) { return x<0? -x:x; }
int S() { 
    return abs((x2-x1)*(y3-y1)-(y2-y1)*(x3-x1))/2;
}
int calc(int x1,int y1,int x2,int y2) {
    return gcd(abs(x2-x1),abs(y2-y1));
}
int main() {
    while(scanf("%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3)==6) {
        if(!x1 && !y1 && !x2 && !y2 && !x3 && !y3) break;
        int b=calc(x1,y1,x2,y2)+calc(x2,y2,x3,y3)+calc(x3,y3,x1,y1);
        printf("%d\n",S()-b/2+1);
    }
    return 0;
}