Problem:


Score : 100 points

Problem Statement

Taro and Jiro will play the following game against each other.

Initially, they are given a sequence a = (a_1, a_2, \ldots, a_N).
Until a becomes empty, the two players perform the following operation alternately, starting from Taro:

Remove the element at the beginning or the end of a. The player earns x points, where x is the removed element.

Let X and Y be Taro's and Jiro's total score at the end of the game, respectively.
Taro tries to maximize X - Y, while Jiro tries to minimize X - Y.

Assuming that the two players play optimally, find the resulting value of X - Y.

Constraints

All values in input are integers.

1 \leq N \leq 3000

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

Print the resulting value of X - Y, assuming that the two players play optimally.

Sample Input 1

4
10 80 90 30

Sample Output 1

10

The game proceeds as follows when the two players play optimally (the element being removed is written bold):

Taro: (10, 80, 90, 30) → (10, 80, 90)

Jiro: (10, 80, 90) → (10, 80)

Taro: (10, 80) → (10)

Jiro: (10) → ()

Here, X = 30 + 80 = 110 and Y = 90 + 10 = 100.

Sample Input 2

3
10 100 10

Sample Output 2

-80

The game proceeds, for example, as follows when the two players play optimally:

Taro: (10, 100, 10) → (100, 10)

Jiro: (100, 10) → (10)

Taro: (10) → ()

Here, X = 10 + 10 = 20 and Y = 100.

Sample Input 3

1
10

Sample Output 3

10

Sample Input 4

10
1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000 1

Sample Output 4

4999999995

The answer may not fit into a 32-bit integer type.

Sample Input 5

6
4 2 9 7 1 5

Sample Output 5

2

The game proceeds, for example, as follows when the two players play optimally:

Taro: (4, 2, 9, 7, 1, 5) → (4, 2, 9, 7, 1)

Jiro: (4, 2, 9, 7, 1) → (2, 9, 7, 1)

Taro: (2, 9, 7, 1) → (2, 9, 7)

Jiro: (2, 9, 7) → (2, 9)

Taro: (2, 9) → (2)

Jiro: (2) → ()

Here, X = 5 + 1 + 9 = 15 and Y = 4 + 7 + 2 = 13.



Sample input:
4
10 80 90 30


Sample output:
10

Submission status: Accepted

Solution code:
N, nums = int(input()), list(map(int, input().split()))

score = [[(-1 if N % 2 == 0 else 1)*x for x in nums]]
for i in range(1, N):
    temp = []
    for j in range(N - i):
        func, sign = (min, -1) if i % 2 == N % 2 else (max, 1)
        temp.append(func(sign*nums[j] + score[-1][j + 1], sign*nums[j + i] + score[-1][j]))
    score.append(temp)

print(score[-1][-1])

Submission status: Accepted

Solution code:
def wczytaj_liste(): 
    wczytana_lista = input()
    lista_znakow = wczytana_lista.split()
    ostateczna_lista = []
    for element in lista_znakow:
        ostateczna_lista.append(int(element))
    return ostateczna_lista 

def japonska_gierka(): 
    N = wczytaj_liste()[0]
    A = wczytaj_liste()
    X = 0
    Y = 0
    wynik = [[None for i in range(N+1)] for j in range(N+1)]
    for dlugosc_kolejki in range(N+1):
        zabrano = N - dlugosc_kolejki
        for lewo in range(zabrano+1): 
            prawo = zabrano - lewo
            if dlugosc_kolejki == 0: 
                wynik[lewo][prawo] = 0 
            else:
                wynik[lewo][prawo] = max(A[lewo]-wynik[lewo+1][prawo], A[-prawo-1]-wynik[lewo][prawo+1])
    print(wynik[0][0])
    
japonska_gierka()

Submission status: Accepted

Solution code:
# input
N = int(input())
As = list(map(int, input().split()))

# Nの偶奇。区間幅がNの偶奇と一致すれば太郎君目線
oe = N % 2

# def print_list(lis):
#   for l in lis:
#     print(l)


# score[i][j][p] : Asのi番目を左端、j番目を右端とした時の最終スコア
#                  p = 0 が太郎君目線、p = 1 が次郎くん目線
score = [[0 for _ in range(N)] for _ in range(N)]

for i in range(N):
  # 初期化
  # 区間長さ1であれば1通りしかない
  if oe == 1:
    score[i][i] = As[i]
  else:
    score[i][i] = -As[i]

# print_list(score)

for k in range(1, N):
  # 配列の幅を広げながら計算を進める
  for i in range(0, N-k):
    # if k == 0:
    #   continue
    j = i + k
    if i == N - 1:
      if (k + 1) % 2 == oe:
        score[i][j] = score[i][j-1] + As[j]
      else:
        score[i][j] = score[i][j-1] - As[j]
    else:
      # print(i,j)
      if (k + 1) % 2 == oe:
        score[i][j] = max(score[i+1][j] + As[i], score[i][j-1] + As[j])
      else:
        score[i][j] = min(score[i+1][j] - As[i], score[i][j-1] - As[j])
    # print_list(score)

print(score[0][N-1])

Submission status: Accepted

Solution code:
# input
N = int(input())
As = list(map(int, input().split()))

# Nの偶奇。区間幅がNの偶奇と一致すれば太郎君目線
oe = N % 2

# def print_list(lis):
#   for l in lis:
#     print(l)


# score[i][j][p] : Asのi番目を左端、j番目を右端とした時の最終スコア
#                  p = 0 が太郎君目線、p = 1 が次郎くん目線
score = [[None for _ in range(N)] for _ in range(N)]

for i in range(N):
  # 初期化
  # 区間長さ1であれば1通りしかない
  if oe == 1:
    score[i][i] = As[i]
  else:
    score[i][i] = -As[i]

# print_list(score)

for k in range(1, N):
  # 配列の幅を広げながら計算を進める
  for i in range(0, N-k):
    # if k == 0:
    #   continue
    j = i + k
    if i == N - 1:
      if (k + 1) % 2 == oe:
        score[i][j] = score[i][j-1] + As[j]
      else:
        score[i][j] = score[i][j-1] - As[j]
    else:
      # print(i,j)
      if (k + 1) % 2 == oe:
        score[i][j] = max(score[i+1][j] + As[i], score[i][j-1] + As[j])
      else:
        score[i][j] = min(score[i+1][j] - As[i], score[i][j-1] - As[j])
    # print_list(score)

print(score[0][N-1])

Submission status: Accepted

Solution code:
n = int(input())
a = [int(x) for x in input().split()]

dp = [[0 for _ in range(n+1)] for _ in range(n+1)]

for i in range(1, n+1):
    for l in range(n):
        if l+i > n:
            continue
        else:
            if i % 2 == n % 2:
                dp[l][l+i] = max(a[l]+dp[l+1][l+i], a[l+i-1]+dp[l][l+i-1])
            else:
                dp[l][l+i] = min(-a[l]+dp[l+1][l+i], -a[l+i-1]+dp[l][l+i-1])

print(dp[0][n])

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int, input().split()))

dp = [[0] * n for _ in range(n)]

for i in range(n):
    dp[i][(n-1)-i] = a[i]

for i in range(n-2, -1, -1):
    for j in range(i+1):
        dp[j][i-j] = max( a[j] - dp[j+1][i-j],
                            a[-(1+i-j)] - dp[j][i-j+1])

print(dp[0][0])

Submission status: Accepted

Solution code:
def deque(n, arr):
    dp = [[0] * n for _ in range(n)]
    for i in range(n):
        dp[i][i] = arr[i]

    for l in range(n-1, -1, -1):
        for r in range(l, n):
            if l == r:
                continue
            L = arr[l] - dp[l+1][r]
            R = arr[r] - dp[l][r-1]
            dp[l][r] = L if L > R else R
    return dp[0][n-1]

n = int(input())
arr = [int(x) for x in input().split()]
print(deque(n, arr))

Submission status: Accepted

Solution code:
def deque(n, arr):
    dp = [[0] * n for _ in range(n)]
    for i in range(n):
        dp[i][i] = arr[i]

    for l in range(n-2, -1, -1):
        for r in range(l+1, n):
            L = arr[l] - dp[l+1][r]
            R = arr[r] - dp[l][r-1]
            dp[l][r] = L if L > R else R
    return dp[0][n-1]

n = int(input())
arr = [int(x) for x in input().split()]
print(deque(n, arr))

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int, input().split()))

dp = [[0]*(n+1) for i in range(n+1) ]
for cnt in range(1, n+1):
    i = 0
    while True:
        j = i + cnt
        if j-1 >= n:
            break
        dp[i][j] = max(a[i] - dp[i+1][j], a[j-1] - dp[i][j-1])
        i += 1
print(dp[0][n])

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    N = r_int()
    A = R()
    dp = [[0]*(N+1) for _ in range(N+1)]
    
    for k in range(1, N+1):
        for j in range(k, N+1):
            # for i in range(j):
            i = j - k
            dp[i][j] = max(A[i] - dp[i+1][j], A[j-1] - dp[i][j-1])
    print(dp[0][N])
    
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
N = int(input())
a = list(map(int, input().split()))

S = [0] * (N+1)
for i in range(N):
    S[i+1] = S[i] + a[i]

dp = [[0 for _ in range(N)] for _ in range(N)]
for i in range(N):
    dp[i][i] = a[i]
for d in range(1, N):
    for i in range(N-d):
        dp[i][i+d] = max(a[i] + S[i+d+1] - S[i+1] - dp[i+1][i+d], a[i+d] + S[i+d] - S[i] - dp[i][i+d-1])
print(dp[0][N-1] * 2 - S[-1])

Submission status: Accepted

Solution code:
N = int(input())
a = [int(c) for c in input().split()]

dp = [[-1] * N for _ in range(N)]
for L in range(N, -1, -1):
    for R in range(L, N):
        if L == R:
            dp[L][R] = a[L]
        else:
            dp[L][R] = max(a[L] - dp[L+1][R], a[R] - dp[L][R-1])


print(dp[0][N-1])

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int,input().split()))
dp = [[0 for j in range(N)] for i in range(N)]
if N%2==0:
    for i in range(N):
        dp[i][i] = -A[i]
    for k in range(1,N):
        for i in range(N-1):
            if i+k<N:
                if k%2==0:
                    dp[i][i+k]= min(-A[i]+dp[i+1][i+k],-A[i+k]+dp[i][i+k-1])
                else:
                    dp[i][i+k] = max(A[i]+dp[i+1][i+k],A[i+k]+dp[i][i+k-1])
else:
    for i in range(N):
        dp[i][i] = A[i]
    for k in range(1,N):
        for i in range(N-1):
            if i+k<N:
                if k%2==0:
                    dp[i][i+k]= max(A[i]+dp[i+1][i+k],A[i+k]+dp[i][i+k-1])
                else:
                    dp[i][i+k] = min(-A[i]+dp[i+1][i+k],-A[i+k]+dp[i][i+k-1])
print(dp[0][N-1])

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int,input().split()))

dp = [[0 for j in range(N+1)] for i in range(N+1)]

for t in range(1,N+1):
    for i in range(N-t+1):
        j = i+t
        if (N-t)%2==0:
            dp[i][j] = max(dp[i+1][j]+A[i],dp[i][j-1]+A[j-1])
        else:
            dp[i][j] = min(dp[i+1][j]-A[i],dp[i][j-1]-A[j-1])

print(dp[0][N])

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int, input().split()))
dp = [[0] * n for _ in range(n)]
for L in range(n - 1, -1, -1):
    for R in range(L, n):
        if L == R:
            dp[L][R] = a[L]
        else:
            dp[L][R] = max(a[L] - dp[L + 1][R], a[R] - dp[L][R - 1])
print(dp[0][n - 1])

Submission status: Accepted

Solution code:
n=int(input())
A=[int(i) for i in input().split()]

dp=[[0 for _ in range(n+1)] for _ in range(n+1)]

for w in range(n):
    for l in range(n-w):
        r=w+l+1
        dp[l][r] = max(-dp[l+1][r]+A[l],-dp[l][r-1]+A[r-1])
print(dp[0][-1])

Submission status: Accepted

Solution code:
import collections
from collections import deque

n=int(input())
a=deque(map(int, input().split()))  
dp=[[0]*(n+1) for i in range(n+1)]

x=0
y=0

for i in range(n-1,-1,-1):
    for j in range(i+1,n+1):
        if (n-(j-i))%2==0:
            dp[i][j]=max(dp[i][j-1]+a[j-1],dp[i+1][j]+a[i])
        else:
            dp[i][j]=min(dp[i][j-1]-a[j-1],dp[i+1][j]-a[i])


print(dp[0][n])

Submission status: Accepted

Solution code:
import sys
# input処理を高速化する
input = sys.stdin.readline
def chmax(a, b):
    if a >= b:
        return a
    return b
def chmin(a, b):
    if a <= b:
        return a
    return b
def main():
    N = int(input())
    a = list(map(int, input().split()))
    # aiからajのX−Yの最大値
    # [左端,右端)=[l,r)からの遷移    
    # 左をとる[l+1,r)、右をとる[l,r−1)の2通り
    dp = [[0] * (N+1)  for _ in range(N+1)]
    for len in range(1, N + 1):
        for i in range(N + 1 - len):
            j = i + len
            if (N - len) % 2 == 0:
                dp[i][j] = chmax(dp[i + 1][j] + a[i], dp[i][j - 1] + a[j - 1])
            else:
                dp[i][j] = chmin(dp[i + 1][j] - a[i], dp[i][j - 1] - a[j - 1])
    print(dp[0][N])
main()

Submission status: Accepted

Solution code:
N = int(input())
a = list(map(int, input().split()))
dp = [[0]*N for _ in range(N)]
if N%2 == 0:
    for i in range(N):
        dp[0][i] = -a[i]
else:
    for i in range(N):
        dp[0][i] = a[i]

for i in range(1, N):
    for j in range(N-i):
        if (N-i)%2 == 1:
            dp[i][j] = max(dp[i-1][j]+a[i+j], dp[i-1][j+1]+a[j])
        else:
            dp[i][j] = min(dp[i-1][j]-a[i+j], dp[i-1][j+1]-a[j])
print(dp[-1][0])

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    A = list(map(int, input().split()))
    c = [0] * N
    for r in reversed(range(N)):
        ar = A[r]
        c[r] = t = ar
        for l in range(r + 1, N):
            c[l] = t = max(ar - c[l], A[l] - t)
    
    print(c[-1])


main()

Submission status: Accepted

Solution code:
def deque(a): # aの先頭or末尾を得点とする A,Bともに最適解の時得点差x-yを求める
    n=len(a)
    # dp[i][j] : a[i],...,a[j-i]の部分文字列でゲームをスタートしたときの最適解
    dp=[[0 for _ in range(n+1)] for _ in range(n+1)] 
    # 初期条件はdp[i][i]=0
    for length in range(1,n+1):
        for i in range(n-length+1):
            j=i+length
            if (n-(j-i))%2==0: # 先手
                dp[i][j]=max( dp[i+1][j] + a[i] , dp[i][j-1] + a[j-1] )
            else: # 後手
                dp[i][j]=min( dp[i+1][j] - a[i] , dp[i][j-1] - a[j-1] )
    print(dp[0][n])

def main():
    int(input())
    a = list(map(int,input().split()))
    deque(a)
if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

from collections import deque

N,*A = map(int,read().split())

B = deque()
for a in A:
    B.append(a)
    while True:
        if len(B) < 3:
            break
        x,y,z = B[-3],B[-2],B[-1]
        if x<=y>=z:
            B.pop()
            B.pop()
            B[-1] = x+z-y
            continue
        break

get = []
while B:
    if B[0] > B[-1]:
        x = B.popleft()
    else:
        x = B.pop()
    get.append(x)

answer = sum(get[::2]) - sum(get[1::2])

print(answer)

Submission status: Accepted

Solution code:
import sys
n = int(input())
a_ls = [int(i) for i in sys.stdin.readline().split()]
memo_ls = [[0 for j in range(n)] for i in range(n)]
for j in range(n):
    for i in range(j,-1,-1):
        if i == j:
            memo_ls[i][j] = a_ls[j]
        else:
            memo_ls[i][j] = max(- memo_ls[i+1][j] + a_ls[i], - memo_ls[i][j-1] + a_ls[j])
print(memo_ls[0][n-1])

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int,input().split()))
dp = [[0 for _ in range(n)] for j in range(n)]
for i in range(n):
    dp[i][i] = a[i]

for dist in range(1,n):
    for i in range(n-dist):
        j = i + dist
        dp[i][j] = max(a[i]-dp[i+1][j],a[j]-dp[i][j-1])

print(dp[0][n-1])

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline

N=int(input())
a = list(map(int, input().split()))

dp = [[None]*(N+1) for _ in range(N+1)]
for i in range(N+1):
    dp[i][i]=0
for len in range(1,N+1):
    for i in range(N-len+1):
        j=i+len
        #print(i,j)
        if (N-len)%2==0:
            dp[i][j] = max(dp[i][j-1]+a[j-1], dp[i+1][j]+a[i])
        else:
            dp[i][j] = min(dp[i][j-1]-a[j-1], dp[i+1][j]-a[i])
#print(dp)
print(dp[0][N])

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int,input().split()))

# dp[i][j]:ai~aj残っているときのX-Yの値。(aiのiは0originとする)
# 太朗君の番なのか次郎君の番なのかは、Nと残っているaの個数から判断する。
dp = [[0 for _ in range(N)] for _ in range(N)]

# 漸化式の右辺にi+1,j-1が出てくるから、
# iは小さいほうから、jは大きいほうから決めていく。
for i in range(N-1,-1,-1):
  for j in range(i,N):
    if (N-(j-i))%2 == 1:
      if i == j:
        dp[i][j] = A[i]
      else:
        dp[i][j] = max(A[i]+dp[i+1][j],A[j]+dp[i][j-1])
    else:
      if i == j:
        dp[i][j] = -A[i]
      else:
        dp[i][j] = min(dp[i+1][j]-A[i],dp[i][j-1]-A[j])
print(dp[0][N-1])

Submission status: Accepted

Solution code:
n=int(input())
a=[int(j) for j in input().split()]
dp=[[0]*(n+1) for i in range(n+1)]

for i in range(n)[::-1]:
    for j in range(i,n):
        if (n+i-j-1)%2==0:
            dp[i][j]=max(dp[i+1][j]+a[i],dp[i][j-1]+a[j])
        else:
            dp[i][j]=min(dp[i+1][j]-a[i],dp[i][j-1]-a[j])
print(dp[0][n-1])