HDU-4969 Just a Joke （数学 - 积分）

Just a Joke

http://acm.hdu.edu.cn/showproblem.php?pid=4969

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Here is just a joke, and do not take it too seriously.

Guizeyanhua is the president of ACMM, and people call him President Guizeyanhua. When Guizeyanhua is walking on the road, everyone eyes on him with admiration. Recently, Guizeyanhua has fallen in love with an unknown girl who runs along the circular race track on the playground every evening. One evening, Guizeyanhua stood in the center of the circular race track and stared the girl soulfully again. But this time he decided to catch up with the girl because of his lovesickness. He rushed to the girl and intended to show her his love heart. However, he could not run too far since he had taken an arrow in the knee. 

Now your task is coming. Given the maximum distance Guizeyanhua can run, you are asked to check whether he can catch up with the girl. Assume that the values of Guizeyanhua's and the girl's velocity are both constants, and Guizeyanhua, the girl, and the center of the circular race track always form a straight line during the process. Note that the girl and Guizeyanhua can be considered as two points.

 

Input

The input begins with a line containing an integer T (T<=100000), which indicates the number of test cases. The following T lines each contain four integers V1, V2, R, and D (0<V1, V2, R, D<=10^9, V1<=V2). V1 is the velocity of the girl. V2 is the velocity of Guizeyanhua. R is the radius of the race track. D is the maximum distance President Guizeyanhua can run.

 

Output

For each case, output "Wake up to code" in a line if Guizeyanhua can catch up with the girl; otherwise output "Why give up treatment" in a line.

 

Sample Input

21 1 1 111904 41076 3561 3613

 

Sample Output

Why give up treatmentWake up to code

一眼就能看出来时微积分，奈何高数远离我太久，化了好久，在离答案只差一个积分时放弃了去吃饭，悔恨不已。。。

推到过程如下：

#include <cstdio>#include <cmath>using namespace std;int main() {    int T,v1,v2,r,d;    scanf("%d",&T);    while(T-->0) {        scanf("%d%d%d%d",&v1,&v2,&r,&d);        printf("%s\n",1.0*v2*r/v1*asin(1.0*v1/v2)<d?"Wake up to code":"Why give up treatment");    }    return 0;}