HackerEarth Ratio - TIE BREAKER problem solution

In this HackerEarthRatio – TIE BREAKER problem solution You are given N ratios in the form of A and B that is represented as A/B. The values of A and B are represented as double data type values. The values of B are incorrect. The actual values of B are B + R. You know the actual sum of all the ratios that is available in variable K.

HackerEarth Ratio – TIE BREAKER problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll n;
double c,t,lo,hi,mid,d[1005],s[1005];
int main()
{
    freopen("samp.txt","r",stdin);
    freopen("sout.txt","w",stdout);
    ll i,j,k=0;
    cin>>n;
    lo=-1000000.0;
    for(i=1;i<=n;i++)
    {
        cin>>d[i]>>s[i];
        lo=max(lo,-1.0*s[i]);
    }
    cin>>t;
    hi=10000000.0;
    k=0;
    while(k<=5000&&lo<=hi)
    {
        mid=(lo+hi)/2.0;
        k++;
        double cur=0.0;
        for(i=1;i<=n;i++)
        {
            cur+=(d[i]/(s[i]+mid+0.0));
        }
        if(cur>=t)
        lo=mid;
        else
        hi=mid;
    }
    cout<<fixed<<setprecision(8)<<mid;
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define eps 0.000001
using namespace std;
int n;
double a[1005],b[1005];
double f(double r)
{
    double val=0.0;
    for(int i=0;i<n;i++)
        val+=a[i]/(b[i]+r);
    return val;
}
int main()
{
    int col;
    cin>>n>>col;
    assert(n>=1 && n<=1000);
    assert(col==2);
    double l=-2000.0,r=2000.0;
    for(int i=0;i<n;i++)
    {
        cin>>a[i]>>b[i];
        assert(a[i]>=1 && a[i]<=1000);
        assert(abs(b[i])<=1000);
        l=max(l,-1.0*b[i]);
    }
    double k,ans=0;l+=1e-7;
    cin>>k;
    assert(k>=1 && k<=1e6);
    while(l<=r)
    {
        double mid=(l+r)/2.0;
        double val=f(mid);
        if(val>=k){ans=mid;l=mid+1e-7;}
        else if(val<k) {r=mid-1e-7;}
    }
    cout<<fixed<<setprecision(7)<<ans<<"n";
    return 0;
}

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