codeforce 9B

最新推荐文章于 2022-05-31 22:37:57 发布

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最新推荐文章于 2022-05-31 22:37:57 发布

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本文介绍了一种算法问题，即如何在多个公交站点中选择最优的下车点以便尽快抵达目的地。考虑了公交车与步行的不同速度，并给出了具体的实现代码。

http://vjudge.net/contest/view.action?cid=47681#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 (x_i, 0)
coordinates of all the stops are different
the minibus drives at a constant speed, equal to v_b
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 (x_u, y_u)
Student can run from a bus stop to the University at a constant speed v_s as long as needed
a distance between two points can be calculated according to the following formula: $\text{[math]}$
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 ≤ v_b, v_s ≤ 1000. The second line contains n non-negative integers in ascending order: coordinates x_i of the bus stop with index i. It is guaranteed that x₁ equals to zero, and x_n ≤ 10⁵. The third line contains the coordinates of the University, integers x_u and y_u, 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

Input

4 5 2
0 2 4 6
4 1

Output

Input

2 1 1
0 100000
100000 100000

Output

Hint

As you know, students are a special sort of people, and minibuses usually do not hurry. That's why you should not be surprised, if Student's speed is higher than the speed of the minibus.

题目大意：在平面直角坐标系中有一点（x，y）从（0,0）出发到该点去，其中在x轴上有一些公交站，可以再00到任意一站之间坐公交，下了站后必须步行到达终点。公交速度和步行速度已给出，求经过几个站后下车，用时最小。第一站为（0,0,）不可下车，若有相同的时间，取距离终点最近的车站下车

纯水题，要注意在double和double比较上的处理，不能用等号。

#include <stdio.h>
#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
double mm(double  x,double y,double a,double b)
{
    return sqrt((x-a)*(x-a)+(y-b)*(y-b));
}
int abs(int x,int y)
{
    if(x>y)
        return x-y;
    return y-x;
}
double a[1005];
double b[1005];
int main()
{
    int n,m,k,x,y;
    while(~scanf("%d%d%d",&k,&n,&m))
    {
        for(int i=0;i<k;i++)
        {
            scanf("%lf",&a[i]);
        }
        scanf("%d%d",&x,&y);
        double minn=999999999;
        for(int i=1;i<k;i++)
        {
            b[i]=(mm(x,y,a[i],0)*1.0/m+a[i]*1.0/n);
            minn=min(b[i],minn);
        }
        int mxxx=99999999;
        for(int i=1;i<k;i++)
             if(fabs(minn-b[i])<1e-10&&abs(i,x)<mxxx)
                mxxx=i;
        printf("%d\n",mxxx+1);
    }
    return 0;
}