poj2954

还是pick定理，直接套用模板，具体请参考上两篇。

 

题目大意：

给你一个三角形，求出三角形内部的整数点的个数。

 

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
//#include <cstring>
#include <cmath>
using namespace std;


struct POINT 
{
	int x,y;
}point[110];

int n;
double getArea()
{
	double sum = 0;
	for (int i = 0; i < n; ++ i)
	{
		sum += (point[i].x   * point[(i + 1) % n].y   - point[i].y  * point[(i + 1) % n].x );
	}
	return fabs(sum/2.0);
}

int Gcd(int a, int b)
{
	if (0 == b)
	{
		return a;
	}
	else
		return Gcd(b, a % b);
}
int getSegmentPoint(POINT p1, POINT p2)
{
	int a = abs(p2.y - p1.y);
	int b = abs(p2.x - p1.x);

	if (a == 0 && b == 0)
	{
		return 0;
	}
	if (a == 0)
	{
		return b - 1;
	}
	if (b == 0)
	{
		return a - 1;
	}
	return Gcd(b, a) - 1;
}
int getPoint()
{
	int ans = n;
	for (int i = 0; i < n; ++ i)
	{
		ans += getSegmentPoint(point[i], point[(i + 1) % n]);
	}
	return ans;
}
int main()
{
	//int cas, j = 1;
	//scanf("%d", &cas);
	n = 3;
	while (1)
	{
		scanf("%d %d %d %d %d %d", &point[0].x, &point[0].y, &point[1].x, &point[1].y, &point[2].x, &point[2].y);
		//point[0].x = point[0].y = 0;
		//for (int i = 1; i <= n; ++ i)
		//{
		//	scanf("%d %d", &point[i].x, &point[i].y);
		//	point[i].x += point[i - 1].x;
		//	point[i].y += point[i - 1].y;
		//	//cin >> point[i].x >> point[i].y;
		//}
		//printf("Scenario #%d:\n", j ++);
		if (point[0].x == 0 && point[0].y == 0 && point[1].x == 0 && point[1].y == 0 && point[2].x == 0 && point[2].y == 0)
		{
			break;
		}
		double Area = getArea();
		int res;
		int PointNum = getPoint();
		res = (int)Area - PointNum / 2 + 1;
		printf("%d\n", res);
	}
	return 0;
}