Kattis - speed (二分)_c - checks post facto kattis

本文链接：https://blog.youkuaiyun.com/dsaghjkye/article/details/79926083

本文介绍了一种通过记录旅程中各段的距离和速度表读数来计算速度表偏差的方法。利用二分搜索确定偏差值，确保实际速度与显示速度之间的正确转换。

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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 s, her true speed is s+c, where c 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 ithof 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 10 5 3 2 2 3 6 3 1	-0.508653377

题意: 你有1个坏的仪表盘，他显示的速度和正确的速度相差 c，现在给你个数n和行驶总时间t ，接下来n行，每行1个行驶距离 d和显示速度s ，现在要求偏差值c 。、

解：比如第1组样例总时间 5 = 4/(c-1) + 4/(c+0) +10/(3+c) ,因为时间 =路程/真实速度；

那么我们可以二分枚举 c 看是否满足要求，首先每个 c+s 必须大于0 ，否则说明 c太小了，

其次将得到的总时间和 t进行比较，若小于 t 说明c太大，若大于t ，说明c太小

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<map>
#include<vector>
#include<set>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
const double eps=1e-9;
int n;
double t;
double d[1100],s[1100];
int aa(double c)
{
    double kk=0;
    for(int i=1;i<=n;i++)
    {
        if(c+s[i]<=0)
            return 0;
        kk+=d[i]/(c+s[i]);
    }
    if(kk>t) return 0;
    else  return 1;
}
int main()
{
    scanf("%d%lf",&n,&t);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf",&d[i],&s[i]);
     double hign=1e8+1;
     double low=-1e8;
     double mid;
     while(hign-low>eps)//因为可能存在误差
     {
         mid=(hign+low)/2;
         if(aa(mid)) hign=mid;
         else low=mid;
     }
     printf("%.6lf\n",hign);
}