Good job! Now that Heidi is able to distinguish between Poisson and uniform distributions, she is in a good position to actually estimate the populations.

Can you help Heidi estimate each village's population?


-----Input-----

Same as the easy version.


-----Output-----

Output one line per village, in the same order as provided in the input, containing your (integer) population estimate.

Your answer is considered correct if it is an integer that falls into the interval $[ \lfloor 0.95 \cdot P \rfloor, \lceil 1.05 \cdot P \rceil ]$, where P is the real population of the village, used to create the distribution (either Poisson or uniform) from which the marmots drew their answers.

Difficulty: interview

Starter code:

Solutions:
v = int(input())
eps = 170

def ans(a):
    a.sort()
    if len(a) % 2 == 0:
        med = a[len(a)//2]
    else:
        med = (a[len(a)//2] + a[len(a)//2 - 1]) // 2

    l = med - med // 2
    r = med + med // 2

    c1 = c2 = 0

    for i in a:
        if i >= l and i <= r:
            c1 += 1
        else:
            c2 += 1

    if abs(c1 - c2) <= eps:
        return (med, "uniform")
    else:
        return (med, "poisson")

    

for i in range(v):
    cur = [int(i) for i in input().split()]
    b = ans(cur)
    if b[1] == "poisson":
        print(b[0])
    else:
        print((max(cur) - min(cur)) // 2)
        
        


def sampleVariance(V):
    X = sum(V) / len(V)
    S = 0.0
    for x in V:
        S += (X-x)**2

    S /= (len(V)-1)
    return (X, S)

#That awkward moment when you realized that variance is sigma^2 but you just took the stat course this semester
for i in range(int(input())):
    V = list(map(int, input().split()))
    X, S = sampleVariance(V)
    v1 = X
    v2 = (2*X) ** 2 / 12

    if abs(v1-S) < abs(v2-S):
        print(int(X))
    else:
        print(max(V)+min(V) // 2)

from statistics import mean
from math import factorial, log, ceil

_lf = []
def logfact(n):
    nonlocal _lf
    CALCLEN = 2200
    if not _lf:
        _lf = [0]*CALCLEN
        for i in range(2, CALCLEN):
            _lf[i] = _lf[i-1] + log(i)
    return _lf[n]

def unif_ml(xs):
    #p1 = mean(xs)
    p2 = max(xs) / 2
    pmle = p2 #max(p1, p2) # probably actually not mle and biased towards 0
    loglh = len(xs) * log(2*pmle + 1)
    return loglh, pmle

def poiss_ml(xs):
    pml = mean(xs)
    loglh = len(xs)*pml + sum([-x*log(pml) + logfact(x) for x in xs])
    return loglh, pml

def solve(xs):
    ul, Pu = unif_ml(xs)
    pl, Pp = poiss_ml(xs)
    #print('dbg:', ul, pl)
    if ul < pl:
        p = 1.01*Pu
    else:
        p = Pp
    print(ceil(p))

def __starting_point():
    v = int(input())
    for i in range(v):
        xs = [int(x) for x in input().split()]
        solve(xs)

__starting_point()