poj2954

来源：互联网发布：java io实例编辑：程序博客网时间：2024/05/02 06:12

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;}elsereturn 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;}