Original proposer: admin

By utilizing for loops, the problem can be solved by managing the following information in the manner of bucket sort:

• \(S_i\), the sum of sizes of birds of kind \(i\)
• \(C_i\), the number of birds of kinds \(i\)

For each kind, the answer can be found as \(S_i/C_i\).
Make sure to print a sufficient number of digits.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n,m;
  cin >> n >> m;
  vector<int> S(m+1,0);
  vector<int> C(m+1,0);
  for(int i=0;i<n;i++){
    int a,b;
    cin >> a >> b;
    S[a]+=b;
    C[a]++;
  }
  for(int i=1;i<=m;i++){
    double oup=S[i];
    oup/=C[i];
    cout << fixed << setprecision(12) << oup << "\n";
  }
  return 0;
}