POJ 1675 Happy Birthday

 

Happy Birthday!

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3352 Accepted: 791

Description

There are three berries on a round birthday cake. You are required to divide the cake into three identical parts such that each part contains exactly one berry. To make it easy, it is assumed that the radius of the berries is 0 and each part of the cake is a sector with 120 degrees. Any line that divides the cake should not go through any berry.

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each case contains exactly 7 integers r, x1, y1, x2, y2, x3 and y3. r is the radius of the cake, (xi, yi) is the coordinates of i-th berry. The center of the cake is at (0, 0) and it's confirmed that all the berries will be on the cake.

Output

For each case, output 'Yes' if there is a valid solution, 'No' otherwise.

Sample Input

22 1 1 -1 1 0 -110 0 9 1 8 -1 8

Sample Output

YesNo

Source

POJ Monthly--2004.06.27 srbga@POJ

 

/*两种方法都可以AC第一种是求圆心到三个顶点构成的线短与X轴正半轴的夹角，假设OA, OB, OC的夹角一次增大，只要夹角差最大值大于等于120即可,注意当有点在原点是输出No其实上述算法的思想简言之就是：遍历任意夹着一条线的另外两条线构成的夹角必须大于120度第二种方法是利用余玄定理求出三条线任意两条之间的夹角的最大值，只要这个最大值大于等于120即可满足条件其实上述算法的思想简言之就是：排除任意夹着一条线的另外两条线构成的夹角小于120度的情况，只是尝试最坏情况下不可行的状态是否存在，而第一种方法是去遍历所有情况寻找不可行状态是否存在因此上述两种方法本质是相同的*///方法1#include <iostream>#define PI 3.141592653589#include <cmath>using namespace std;void swap(double &v1, double &v2){ double temp = v1; v1 = v2; v2 = temp;}double getAngle(double x, double y){ double l = sqrt(x * x + y * y); double res;//由于acos算法出的角度范围为0 ～ 180 度因此需要区分对待正负半轴//Y轴正半轴 if(y >= 0) res = acos(x / l); //Y轴负半轴else res = 2 * PI - acos(x / l); return res / PI * 180;}int main(){ int caseNum; int r, x1, y1, x2, y2, x3, y3; scanf("%d", &caseNum); while(caseNum--) { scanf("%d%d%d%d%d%d%d", &r, &x1, &y1, &x2, &y2, &x3, &y3); bool can = true; if((x1 == 0 && y1 == 0) || (x2 == 0 && y2 == 0) || (x3 == 0 && y3 == 0)) can = false; if(can) {//按升序生成三个夹角 double a1 = getAngle(x1, y1); double a2 = getAngle(x2, y2); double a3 = getAngle(x3, y3); if(a1 > a2) swap(a1, a2); if(a2 > a3) swap(a2, a3); if(a1 > a2) swap(a1, a2); //第二条线包含在第一和第三条线中间 if(a3 - a1 <= 120) can =false;//第一条线包含在第二和第三条线中间 if(360 - a3 + a2 <= 120) can = false;//第三条线包含在第一和第二条线中间 if(360 - a2 + a1 <= 120) can = false; } if(can) printf("Yes/n"); else printf("No/n"); } return 0;}//方法2#include <iostream>#define PI 3.141592653589#include <cmath>using namespace std;double maxv(double v1, double v2){ if(v1 >= v2) return v1; else return v2;}double getAngle(double x1, double y1, double x2, double y2){ double up = x1 * x2 + y1 * y2; double down = sqrt(x1 * x1 + y1 * y1) * sqrt(x2 * x2 + y2 * y2); double angle = acos(up / down); angle = angle / PI * 180; return angle;}int main(){ int caseNum; int r, x1, y1, x2, y2, x3, y3; scanf("%d", &caseNum); while(caseNum--) { scanf("%d%d%d%d%d%d%d", &r, &x1, &y1, &x2, &y2, &x3, &y3); bool can = true; if((x1 == 0 && y1 == 0) || (x2 == 0 && y2 == 0) || (x3 == 0 && y3 == 0)) can = false; if(can) { double maxAngle = getAngle(x1, y1, x2, y2); maxAngle = maxv(maxAngle, getAngle(x1, y1, x3, y3)); maxAngle = maxv(maxAngle, getAngle(x2, y2, x3, y3)); if(maxAngle < 120) can = false; } if(can) printf("Yes/n"); else printf("No/n"); } return 0;}