Weakness and Poorness (Codeforces 578C)
Tags: Others, ternary search

You are given a sequence of n integers a1, a2, ..., an. Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.The poorness of a segment is defined as the absolute value of sum of the elements of segment.

Input: The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).

Output: Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.

Note: For the first case, the optimal value of x is 2 so the sequence becomes  - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case. For the second sample the optimal value of x is 2.5 so the sequence becomes  - 1.5,  - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.

Sample Inputs:
3
1 2 3

4
1 2 3 4

10
1 10 2 9 3 8 4 7 5 6

Sample Outputs:
1.000000000000000

2.000000000000000

4.500000000000000

Solution (Python):
def f(x):
    bi = 0
    ci = 0
    cnt1 = -10 ** 10
    minsum1 = 10 ** 10
    cnt2 = -10 ** 10
    minsum2 = 10 ** 10
    for i in range(len(A)):
        if bi - minsum1 > cnt1:
            cnt1 = bi - minsum1
        if bi <= minsum1:
            minsum1 = bi
        if ci - minsum2 > cnt2:
            cnt2 = ci - minsum2
        if ci <= minsum2:
            minsum2 = ci
        bi += A[i] - x
        ci += x - A[i]
    if bi - minsum1 > cnt1:
        cnt1 = bi - minsum1
    if bi <= minsum1:
        minsum1 = bi
    if ci - minsum2 > cnt2:
        cnt2 = ci - minsum2
    if ci <= minsum2:
        minsum2 = ci
    return max(cnt1, cnt2, 0)
n = int(input())
A = list(map(int, input().split()))
l = min(A)
r = max(A)
for _ in range(70):
    p = l + (r - l) / 3
    q = r - (r - l) / 3
    x = f(p)
    y = f(q)
    if x > y:
        l = p
    elif y > x:
        r = q
    else:
        l = p
        r = q
print(f(l))

Solution (Python):
n = int(input())
A = list(map(int, input().split()))

def f(A, x, sign=1):
    cur = 0
    mn = 0
    ans = -1 << 32
    if sign == 1:
        for y in A:
            cur += y - x
            ans = max(ans, cur - mn)
            mn = min(cur, mn)
    else:
        for y in A:
            cur -= y - x
            ans = max(ans, cur - mn)
            mn = min(cur, mn)
    return ans
(lo, hi) = (-10000, 10000)
nt = 100
while nt:
    nt -= 1
    mid = (lo + hi) / 2
    X = f(A, mid, sign=1)
    Y = f(A, mid, sign=-1)
    if X > Y:
        lo = mid
    else:
        hi = mid
print(abs(f(A, lo)))

Solution (Python):
import sys
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split()]
eps = 1e-12

def f(x):
    mx = a[0] - x
    tsmx = 0.0
    mn = a[0] - x
    tsmn = 0.0
    for ai in a:
        tsmx = max(tsmx + ai - x, ai - x)
        mx = max(tsmx, mx)
        tsmn = min(tsmn + ai - x, ai - x)
        mn = min(tsmn, mn)
    return (abs(mx), abs(mn))
l = min(a)
r = max(a)
(f1, f2) = f(l)
for i in range(0, 90):
    m = (l + r) / 2
    (f1, f2) = f(m)
    if f1 > f2:
        l = m
    else:
        r = m
(A, B) = f(l)
print(min(A, B))