The average of ~n~ numbers ~a_1, \dots, a_n~ is given by ~\frac 1 n \sum a_i~. Knowing this, implement what is required in the problem statement, and pay special attention to print the result with an absolute error no greater than ~10^{-6}~.

In C/C++: printf("%.6f\n", ans)
In C++: std::cout << std::fixed << std::setprecision(6) << ans << "\n"
In Java: System.out.printf("%.6f\n", ans)
In Python: print(f'This converts to {ans:.2f}')

Pseudocode:

read A, B, C, D

print (A + B) / 2.0
print (A + C) / 2.0
print (A + D) / 2.0
print (B + C) / 2.0
print (B + D) / 2.0
print (C + D) / 2.0
print (A + B + C) / 3.0
print (A + B + D) / 3.0
print (A + C + D) / 3.0
print (B + C + D) / 3.0
print (A + B + C + D) / 4.0

Time complexity: ~\mathcal{O}(1)~