HDU3086-追及问题-解一元二次方程组

Need for Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 759    Accepted Submission(s): 214

Problem Description

Jiaoshou is a fan of “Need for Speed”, a famous game which most people have played. Recently, Jiaoshou had cleared “Need for Speed Undercover”. In this game, Jiaoshou acted as a spy. To prove that he is not a spy, Jiaoshou must do some evil things frequently. So lots of police cars often ran after him. In order to escape from the chase, Jiaoshou need to accelerate his Bugatti Veyron. 

To simplify this problem, we just assume that only one police car ran after Jiaoshou. And Bugatti Veyron’s speed is vx. The police car’s speed is vy. v is a function of first degree or a constant, as the form of a*m+b. In this form, m means the minute that the car has run.
Now Jiaoshou is S meters ahead. Jiaoshou like to know if he could escape from the chase. He want you to write a program to solve this problem.

Input

The input contains several tests. The first line of the input is a single integer T (T<1000) which is the number of test cases. Then T test cases follow.
Each test case contains three lines. The first line contains a single real number S (0<S<=10^6) , as the problem has described. The second line contains two real numbers ax, bx, indicating Bugatti Veyron’s speed is ax*m+bx. The last line also contains two real numbers ay, by, indicating the police car’s speed is ay*m+by. All numbers are range from -1000 to 1000, except S.

Output

If Jiaoshou can escape from the chase, please output “Good driver,Jiaoshou!”. Otherwise, please output the real number M rounded to three digits after the decimal point , in the form of “Oh,God!Jiaoshou will be catched after M minutes.” 

Sample Input

21000 100 2010000 10 0

Sample Output

Oh,God!Jiaoshou will be catched after 10.000 minutes.Good driver,Jiaoshou!

Author

Water Problem SecKill Expert

Source

HDU 2nd “Vegetable-Birds Cup” Programming Open Contest

 

思路:

很明确的思路,A的速度为ax*m+bx,B的速度为ay*m+by,A在B之前dist远处,问B能否追上A,如果能追上,那么经过多少分钟才能

追上?建立时间-速度的二维坐标,那么速度对时间的积分即为距离(物理公式:s=vt),ax+b的定积分为ax^2/2+bx,所以,化简可以得到一

个一元二次方程组,接下来的就是解方程组了.

/*Tdistax bxay by追及问题:(ay-ax)/2*m^2 + (by-bx)*m - dist = 0 判断是否有解*/#include <iostream>#include <math.h>using namespace std;const double esp = 0.0000000000001;bool isExistSolution;double ans1,ans2;void solve_equation(double a,double b,double c) //ax^2+bx+c=0{if (fabs(a) < esp){if (fabs(b) > 0){isExistSolution = true;ans1 = ans2 = -c / b;// 分母不能为0!!}elseisExistSolution = false;return;}double deltar = b*b - 4.0*a*c;if (deltar < 0)isExistSolution = false;else{isExistSolution = true;deltar = sqrt(deltar)/(2.0*a);b = (-b)/(2.0*a);ans1 = b + deltar;ans2 = b - deltar;}}int main(){#ifndef ONLINE_JUDGEfreopen("2.txt","r",stdin);#endifint T;double dist, ax, bx, ay, by;cin >> T;while (T--){cin >> dist >> ax >> bx >> ay >> by;solve_equation(0.5*(ay-ax),by-bx,-dist);if (!isExistSolution)puts("Good driver,Jiaoshou!");else{if (ans1 > 0)printf("Oh,God!Jiaoshou will be catched after %.3lf minutes.\n",ans1);else if (ans2 > 0)printf("Oh,God!Jiaoshou will be catched after %.3lf minutes.\n",ans2);elseputs("Good driver,Jiaoshou!");}}return 0;}