Intercity Travelling (Codeforces 1009E)
Tags: Math, probabilities

Leha is planning his journey from Moscow to Saratov. He hates trains, so he has decided to get from one city to another by car.The path from Moscow to Saratov can be represented as a straight line (well, it's not that straight in reality, but in this problem we will consider it to be straight), and the distance between Moscow and Saratov is $$$n$$$ km. Let's say that Moscow is situated at the point with coordinate $$$0$$$ km, and Saratov — at coordinate $$$n$$$ km.Driving for a long time may be really difficult. Formally, if Leha has already covered $$$i$$$ kilometers since he stopped to have a rest, he considers the difficulty of covering $$$(i + 1)$$$-th kilometer as $$$a_{i + 1}$$$. It is guaranteed that for every $$$i \in [1, n - 1]$$$ $$$a_i \le a_{i + 1}$$$. The difficulty of the journey is denoted as the sum of difficulties of each kilometer in the journey.Fortunately, there may be some rest sites between Moscow and Saratov. Every integer point from $$$1$$$ to $$$n - 1$$$ may contain a rest site. When Leha enters a rest site, he may have a rest, and the next kilometer will have difficulty $$$a_1$$$, the kilometer after it — difficulty $$$a_2$$$, and so on.For example, if $$$n = 5$$$ and there is a rest site in coordinate $$$2$$$, the difficulty of journey will be $$$2a_1 + 2a_2 + a_3$$$: the first kilometer will have difficulty $$$a_1$$$, the second one — $$$a_2$$$, then Leha will have a rest, and the third kilometer will have difficulty $$$a_1$$$, the fourth — $$$a_2$$$, and the last one — $$$a_3$$$. Another example: if $$$n = 7$$$ and there are rest sites in coordinates $$$1$$$ and $$$5$$$, the difficulty of Leha's journey is $$$3a_1 + 2a_2 + a_3 + a_4$$$.Leha doesn't know which integer points contain rest sites. So he has to consider every possible situation. Obviously, there are $$$2^{n - 1}$$$ different distributions of rest sites (two distributions are different if there exists some point $$$x$$$ such that it contains a rest site in exactly one of these distributions). Leha considers all these distributions to be equiprobable. He wants to calculate $$$p$$$ — the expected value of difficulty of his journey.Obviously, $$$p \cdot 2^{n - 1}$$$ is an integer number. You have to calculate it modulo $$$998244353$$$.

Input: The first line contains one number $$$n$$$ ($$$1 \le n \le 10^6$$$) — the distance from Moscow to Saratov.The second line contains $$$n$$$ integer numbers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_1 \le a_2 \le \dots \le a_n \le 10^6$$$), where $$$a_i$$$ is the difficulty of $$$i$$$-th kilometer after Leha has rested.

Output: Print one number — $$$p \cdot 2^{n - 1}$$$, taken modulo $$$998244353$$$.


Sample Inputs:
2
1 2

4
1 3 3 7

Sample Outputs:
5

60

Solution (Python):
(i, m) = (input, 998244353)
i()
a = map(int, i().split())
(r, c) = [next(a)] * 2
for j in a:
    (r, c) = ((2 * r + c + j) % m, (2 * c + j) % m)
print(r)

Solution (Python):
""""""
from __future__ import division, print_function
import cmath
import itertools
import math
import operator as op
import sys
from atexit import register
from bisect import bisect_left, bisect_right
if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream
if sys.version_info[0] < 3:

    class dict(dict):
        """"""

        def items(self):
            """"""
            return dict.iteritems(self)

        def keys(self):
            """"""
            return dict.iterkeys(self)

        def values(self):
            """"""
            return dict.itervalues(self)
    input = raw_input
    range = xrange
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

def sync_with_stdio(sync=True):
    """"""
    global input, flush
    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda : sys.stdin.readline().rstrip('\r\n')
        sys.stdout = stream()
        register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))

def gcd(x, y):
    """"""
    while y:
        (x, y) = (y, x % y)
    return x

def main():
    input()
    a = map(int, input().split(' '))
    (ediff, count) = [next(a)] * 2
    for ai in a:
        ediff = (2 * ediff + count + ai) % 998244353
        count = (2 * count + ai) % 998244353
    print(ediff)
if __name__ == '__main__':
    sync_with_stdio(False)
    main()

Solution (Python):
def main():
    input()
    (s, *aa) = map(int, input().split())
    y = s
    for a in aa:
        s = (s * 2 + y + a) % 998244353
        y = (y * 2 + a) % 998244353
    print(s)
if __name__ == '__main__':
    main()