HackerEarth Ratio – TIE BREAKER problem solution YASH PAL, 31 July 202415 February 2026 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 intll 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 0.000001using 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;} coding problems solutions HackerEarth HackerEarth