World Finals 2017Need for Speed

Need for Speed

 

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 cc. The journey consisted of nn segments. In the ithith segment she traveled a distance of didi and the speedometer read sisi for the entire segment. This whole journey took time tt. Help Sheila by computing cc.

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 nn (1≤n≤10001≤n≤1000), the number of sections in Sheila’s journey, and tt(1≤t≤1061≤t≤106), the total time. This is followed by nn lines, each describing one segment of Sheila’s journey. The ithith of these lines contains two integers didi (1≤di≤10001≤di≤1000) and sisi (|si|≤1000|si|≤1000), the distance and speedometer reading for the ithith segment of the journey. Time is specified in hours, distance in miles, and speed in miles per hour.

Output

Display the constant cc in miles per hour. Your answer should have an absolute or relative error of less than 10−610−6.

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

wa了好久了，搞不懂二分范围，以为-1000到1e6+1000就够了，我一开大范围反而得不到答案

通过读题列出通项公式，很容易想到二分操作

#include<bits/stdc++.h>
#define eps 1e-8
using namespace std;
typedef long long ll;
double a[1005],b[1005];
int n,m;
double solve(){
    double l=-1<<30,r=1<<30,mid;
    for(int i=0;i<n;i++)
    l=max(l,-b[i]);
    for(int i=0;i<10000;i++){
        mid=(l+r)/2;
        double sum=0.0;
        for(int j=0;j<n;j++){
            sum+=a[j]/(mid+b[j]);
        }
        if(sum-m>0) l=mid;
        else r=mid;
    }
    return r;
}
int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++)
    cin>>a[i]>>b[i];
    double ans=solve();
    printf("%.8lfn",ans);
}