Kattis - speed

最新推荐文章于 2020-02-07 22:43:31 发布

原创最新推荐文章于 2020-02-07 22:43:31 发布 · 283 阅读

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本文介绍了一种通过已知行程和时间来计算车辆速度计误差的方法。利用二分查找算法确定速度计显示值与实际速度之间的常数误差。文章提供了一个具体的实现案例。

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Sheila is a student and she drives a typical student car: it is old, slow, rusty, and falling apart. Recently, the needle on the speedometer fell off. She glued it back on, but she might have placed it at the wrong angle. Thus, when the speedometer reads

ss, her true speed is

s+cs+c, where

cc is an unknown constant (possibly negative).

Sheila made a careful record of a recent journey and wants to use this to compute c. The journey consisted of n segments. In the ith segment she traveled a distance of di and the speedometer read si for the entire segment. This whole journey took time t. Help Sheila by computing c.

Note that while Sheila’s speedometer might have negative readings, her true speed was greater than zero for each segment of the journey.

Input

The first line of input contains two integers n (1≤n≤1000), the number of sections in Sheila’s journey, and t (1≤t≤106), the total time. This is followed by n lines, each describing one segment of Sheila’s journey. The ith of these lines contains two integers di (1≤di≤1000) and si (|si|≤1000), the distance and speedometer reading for the ith segment of the journey. Time is specified in hours, distance in miles, and speed in miles per hour.

Output

Display the constant c in miles per hour. Your answer should have an absolute or relative error of less than 10−6.

Sample Input 1	Sample Output 1
3 5 4 -1 4 0 10 3	3.000000000

Sample Input 2	Sample Output 2
4 105 32 23 63 1	-0.508653377

Sample Input 2

Sample Output 2

4 105 32 23 63 1

-0.508653377

题意:

给出n段车速，m为n段车速的正确平均速度，在n段车速中存在误差，正确为s+c，求c。

思路:

二分查找，double型小数的二分。需要控制精度。还有一个坑点在于每一段的速度只能是正数，也就是说s+c>0，这里需要特判。

#include<stdio.h>
const double ppx=0.0000000001;
int n;
double m,s[1010],t[1010];
int main()
{
    scanf("%d%lf",&n,&m);
    double sum=0,pans;
    for(int i=0; i<n; i++)
    {
        scanf("%lf%lf",&s[i],&t[i]);
        sum+=s[i];
    }
    pans=sum/m;
    int flag;
    double l=-1e8,r=1e8,pan,ans;
    while(l<=r)
    {
        flag=0;
        pan=(l+r)/2;
        ans=0;
        for(int i=0; i<n; i++)
        {
            ans=ans+s[i]/(t[i]+pan);
            if(t[i]+pan<=0)
            {
                flag=1;
                break;
            }
        }
        if(flag)
        {
            l=pan+ppx;
            continue;
        }
        else if(ans-m>ppx)
            l=pan+ppx;
        else
            r=pan-ppx;
    }printf("%.9lf\n",pan);
}