bnu10805矩形神码的

矩形神码的

Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer IO format:  %lld      Java class name:  Main Special Judge Prev  Submit  Status  Statistics  Discuss   Next

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我们都知道，矩形是由两条对角线的，没错吧？（谜之声：这不是显然么！）这两条线的长度也是相等的，没错吧？（谜之声：这不废话么！）然后我们给定一条对角线的起始点和终止点的坐标，然后给定另一个对角线和他的夹角，是不是就能得到两个面积相等的矩形？（谜之声：呃，貌似好像或许应该可能maybe perhaps probably possibly是对的？） 现在我需要你求出这个矩形的面积。

Input

第一行，一个整数T（0<T<=10000），表示数据组数。 接下来T行，每行5个浮点数。分别是x1,y1,x2,y2,theta。表示一条对角线的起始点和终止点的坐标，以及另一条对角线与他的夹角。坐标的绝对值范围均在10 4以内，夹角范围(0,90]。

Output

对于每一组数据，输出一行，表示面积，精确到小数点后六位。 绝对误差或者相对误差在0.0001内均算通过。

Sample Input

2 1.0 1.0 -1.0 -1.0 90.0 3.0 2.0 2.5 9.99 36.00

Sample Output

4.000000 18.835608

Hint

pi请取值acos(-1.0)或者3.1415926535898

#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #define pi 3.1415926535898 using namespace std; int main() { double x1,y1,x2,y2; double t; int n; scanf("%d",&n); while(n--) { scanf("%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&t); double d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))/2; double a=t/180*pi; double b=pi-a; double s=d*d*sin(a)+d*d*sin(b);//三角形面积公式1/2bcsina printf("%.6lf\n",s); } return 0; }