Гра в картки (Розбір)

Let the array $m$ contain the numbers written on the cards. We initialize pointers i = 0 at the beginning of the array and j = n - 1 at the end of the array.

On each turn, a player takes the card with the maximum value max (m[i], m[j]), after which the card is removed from the table (perform the operation i++ if the card $m[i]$ is chosen, and j-- if the card $m[j]$ is chosen). For each player, we keep track of the sum of the selected cards in two variables. The process continues until all cards are removed from the table, that is, until the pointers $i$ and $j$ meet.

Example

The game will proceed as follows.

Algorithm realization

Declare the arrays. The sum of the numbers on the cards collected by Huseyn and Yaroslav will be stored in $res[0]$ and $res[1]$, respectively.

int m[10001], res[2];

Read the input data.

scanf("%d", &n);
for (i = 0; i < n; i++)
  scanf("%d", &m[i]);

Set the pointers i = 0 and j = n - 1.

i = 0; j = n - 1;

The players take turns, making $n$ moves in total. Huseyn takes turns for even $k$, and Yaroslav takes turns for odd $k$. Accordingly, Huseyn's sum will accumulate in $res[0]$, and Yaroslav's sum will accumulate in $res[1]$.

for (k = 0; k < n; k++)
{

On each turn, the player chooses the card with the maximum value max(m[i], m[j]).

  if (m[i] > m[j])
    res[k % 2] += m[i++];
  else
    res[k % 2] += m[j--]; ;
}

Print the answer.

printf("%d %d\n", res[0], res[1]);

Python realization

Read the input data.

n = int(input())
m = list(map(int, input().split()))

Set the pointers i = 0 and j = n - 1.

i, j = 0, n – 1

The sum of the numbers on the cards collected by Huseyn and Yaroslav will be stored in $res[0]$ and $res[1]$, respectively.

res = [0, 0]

The players take turns, making $n$ moves in total. Huseyn takes turns for even $k$, and Yaroslav takes turns for odd $k$. Accordingly, Huseyn's sum will accumulate in $res[0]$, and Yaroslav's sum will accumulate in $res[1]$.

for k in range(n):

On each turn, the player chooses the card with the maximum value max(m[i], m[j]).

  if m[i] > m[j]:
    res[k % 2] += m[i]
    i += 1
  else:
    res[k % 2] += m[j]
    j -= 1

Print the answer.

print(res[0], res[1])