B. Running Student
题目连接：
http://www.codeforces.com/contest/9/problem/B
Description
And again a misfortune fell on Poor Student. He is being late for an exam.
Having rushed to a bus stop that is in point (0, 0), he got on a minibus and they drove along a straight line, parallel to axis OX, in the direction of increasing x.
Poor Student knows the following:
during one run the minibus makes n stops, the i-th stop is in point (xi, 0)
coordinates of all the stops are different
the minibus drives at a constant speed, equal to vb
it can be assumed the passengers get on and off the minibus at a bus stop momentarily
Student can get off the minibus only at a bus stop
Student will have to get off the minibus at a terminal stop, if he does not get off earlier
the University, where the exam will be held, is in point (xu, yu)
Student can run from a bus stop to the University at a constant speed vs as long as needed
a distance between two points can be calculated according to the following formula:
Student is already on the minibus, so, he cannot get off at the first bus stop
Poor Student wants to get to the University as soon as possible. Help him to choose the bus stop, where he should get off. If such bus stops are multiple, choose the bus stop closest to the University.
Input
The first line contains three integer numbers: 2 ≤ n ≤ 100, 1 ≤ vb, vs ≤ 1000. The second line contains n non-negative integers in ascending order: coordinates xi of the bus stop with index i. It is guaranteed that x1 equals to zero, and xn ≤ 105. The third line contains the coordinates of the University, integers xu and yu, not exceeding 105 in absolute value.
Output
In the only line output the answer to the problem — index of the optimum bus stop.
Sample Input
4 5 2
0 2 4 6
4 1
Sample Output
3
Hint
题意
有一个小朋友，考试要迟到了，所以必须尽快的赶到学校。
他跑步的速度是v2,公交车的速度是v1，现在他可以选择从某一站下车，然后飞奔过去
现在问你应该从哪一站下车。
不允许从第一站下车啦，因为他就是从那儿上车的。
题解：
噜噜噜，水题啦，直接暴力枚举从哪儿下车就好了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int a[maxn];
double dis(double x1,double y1,double x2)
{
return sqrt((x1-x2)*(x1-x2)+y1*y1);
}
int main()
{
int n;
double v1,v2;
scanf("%d%lf%lf",&n,&v1,&v2);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
double x,y;cin>>x>>y;
double ans2 = dis(x,y,a[2])/v2+a[2]/v1;
int ans1 = 2;
for(int i=3;i<=n;i++)
{
double tmp = a[i]/v1+dis(x,y,a[i])/v2;
if(tmp<=ans2)ans2=tmp,ans1=i;
}
cout<<ans1<<endl;
}