发布日期:2022-02-08 04:20:46 浏览次数:1 分类:技术文章

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Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 16   Accepted Submission(s) : 7
Problem Description
The recreation center of WHU ACM Team has indoor billiards, Ping Pang, chess and bridge, toxophily, deluxe ballrooms KTV rooms, fishing, climbing, and so on.<br>We all like toxophily.<br><br>Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?<br><br>Now given the object's coordinates, please calculate the angle between the arrow and x-axis at Bob's point. Assume that g=9.8N/m. <br>

The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case is on a separated line, and it consists three floating point numbers: x, y, v. x and y indicate the coordinate of the fruit. v is the arrow's exit speed.<br>Technical Specification<br><br>1. T ≤ 100.<br>2. 0 ≤ x, y, v ≤ 10000. <br>

For each test case, output the smallest answer rounded to six fractional digits on a separated line.<br>Output "-1", if there's no possible answer.<br><br>Please use radian as unit. <br>




判断一下x1,x2是否在[          0       ,      pi       ]之间然后选一个合法的较小值即可 

using namespace std;#define g 9.8#define pi acos(-1.0)/2.0int main(){ int T; double x,y,v; scanf("%d",&T); while(T--) { scanf("%lf%lf%lf",&x,&y,&v); double a=g*x*x; double b=-2.0*x*v*v; double c=2.0*y*v*v+g*x*x; double d=b*b-4.0*a*c; if(d<0) { printf("-1\n"); continue; } d=sqrt(d); double x1=atan((-b+d)/(2.0*a)),x2=atan((-b-d)/(2.0*a)); if(x1<0||x1>pi&&x2<0||x2>pi)printf("-1\n"); else if(x1<0||x1>pi)printf("%.6lf\n",x2); else if(x2<0||x2>pi)printf("%.6lf\n",x1); else printf("%.6lf\n",x1

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[***.202.152.39]2023年09月18日 02时31分46秒