Kattis - 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 ithithof 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

3 5
4 -1
4 0
10 3

Sample Output 1

3.000000000

Sample Input 2
4 10
5 3
2 2
3 6
3 1

Sample Output 2

-0.508653377

题意：给出一个方程  s1/(v1+c)+s2/(v2+c)+s3/(v3+c)+........=t; si，vi 和 t都是已知的

思路：由于方程是具有单调性的而且好像没有其他解法，所以可以用二分来求解。

#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define maxn 1005
#define ll long long
#define inf 0x3f3f3f
using namespace std;
double s[maxn],v[maxn];
int n;double t;
bool judge(double x){
    double sum=0;
    for(int i=0;i<n;i++){
        if(v[i]+x<0)return true;
        sum+=(s[i]/(v[i]+x));
    }
    if(sum>t)return true;
    else
        return false;
}
int main(){
    while(cin>>n>>t){
        double l=-inf,r=inf,mid;
        for(int i=0;i<n;i++){
            cin>>s[i]>>v[i];
        }
        while(r-l>0.000000001){//题目要求误差在1e-6以内。
            mid=(l+r)/2;
            if(judge(mid))l=mid;
            else r=mid;
        }
        printf("%.9lf\n",mid);
    }
}