POJ2954Triangle【pick公式】

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Triangle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5472 Accepted: 2370

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 10 0 5 0 0 50 0 0 0 0 0

Sample Output

06

题意：和poj1265一样只不过本题为三角形做poj1265不知道两求两点之间的点数用了枚举加点在线段上的判断之后看了别人的代码发现可以直接求两点的gcd获得除起点外在该线段上的点数所以用到这道题上刚好。

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>using namespace std;int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){int x1,x2,x3,y1,y2,y3,i,j,k;while(scanf("%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3),x1||x2||x3||y1||y2||y3){int a=gcd(abs(x2-x1),abs(y2-y1));int b=gcd(abs(x3-x2),abs(y3-y2));int c=gcd(abs(x3-x1),abs(y3-y1));int s=(x3-x2)*(y1-y2)-(y3-y2)*(x1-x2);printf("%d\n",(abs(s)+2-a-b-c)/2);}return 0;}