Idea: SlavicG

Create a variable max, a variable sum and a variable pos. Set $max = 0$, $sum = 0$, $pos= 0$. Iterate through the array and add $arr[I]$ to $sum$. If $sum > max$ then save $i$ in the variable $pos = i$, and set max to the current $sum$. After you are done iterating print $pos$.

Link to solution: https://ideone.com/wZWJBW

Idea: mesanu

The following greedy solutions always works: Sort the array $arr$, and then iterate until you find the minimum even number. If at the end of the loop it doesn’t exist then we have no answer and the solution is $-1$. Because we can not create an even number without any even digits. If there is an even element, save it’s index to a variable.Then print the array decreasing order (from $n$ to $1$)without the minimum even digit(the element at the saved index). After printing the array, print the minimum even digit so the number will be even.

Link to solution: https://ideone.com/fyDKc2

Idea: SlavicG This problem can be solved using dynamic programming. Set an array $dp[n+1]$ and all the elements are equal to $10^9$ and set $dp[1]$ = 0; iterate through the array from $2$ to $n$ and for each $dp[i] = min(dp[i], dp[j]+abs(arr[i]-arr[j]))$ where $(i-k \leq j < i)$ and $j$ indicates one of the last $k$ bus stops. After you are done iterating print $dp[n+1]$.

Link to solution: https://ideone.com/jDkZE0

Idea: mesanu

There are $2$ cases: If $k$ is even Bob makes the last move and he can always insert a prime number such that the $gcd$ of the array is $1$.

If $k$ is odd then Alice makes the last move. Alice wins if we can remove an element from the starting array such that the $gcd$ of $arr$ is bigger than 1. To find this out we can use the fundamental theory of arithmetic which states that any integer can be expressed as a product of prime numbers.For each element in the array we find what primes divide it by iterating through all the primes under 100(since it is given that $a_i \leq 100$). If any prime divides more than $n-1$ elements Alice wins since if Bob introduces a number Alice can delete that number in the next move. If no prime divides more than $n-1$ elements then bob wins.

Link to solution: https://ideone.com/nmWCRz

Idea: SlavicG Another solution for when $k$ is odd is with prefix and suffix $gcd$. Iterrate over all $a_i$ from $1$ to $n$ and set $prefixgcd[i] = gcd(prefixgcd[i-1],prefixgcd[i])$. Then do the same for $suffixgcd$, but in reverse order: from $n$ to $1$ and set every element $suffixgcd[i] = gcd(suffixgcd[i+1],suffixgcd[i])$. Be careful at $prefixgcd_1$ and $suffixgcd_n$, You should just attribute them the value of the element at the index. After that we just iterate over all the array and check if $gcd(prefixgcd[i-1], suffixgcd_[i+1]) > 1$ ( again, be careful at $a_1$ and $a_n$). If this is true for at least one element then the answer is $"YES"$. But if we can not find such i then the answer is $"NO"$.

Idea: mesanu

We create an array $sum$ where $sum[i]$ is the sum of all the elements in array $a$ with index smaller than one. Set $sum[0]$ = 0. We have $sum[i] = sum[i-1]+a[i-1]$. Now we can use binary search to find the smallest sum bigger than $b[i] then print the index where the smallest sum bigger than b[i] is.

Link to solution: https://ideone.com/1DCK0b