Cf:e. price maximization [sort + take mod + double pointer + pair]

analysis

First, the remainder of each number is obtained , The integer part can be removed first
Then sort the remainder
And then the two pointers end to end , If it adds up, it can be greater than or equal to k Words , Then integer part +1, Then both pointers move to the middle
Otherwise , The element that the left pointer currently refers to cannot find a suitable match , So move to the right , Then continue to judge

ac code

for _ in range(int(input())):
    n, k = list(map(int, input().split()))
    a = list(map(int, input().split()))
 
    sumA = sum([num // k for num in a])
    mod = [num % k for num in a]
    mod.sort()
    # print(new_a)
    bonus = 0
 
    s, e = 0, n - 1
    # greedy + double points
    while s < e:
        # gather
        if mod[s] + mod[e] >= k:
            bonus += 1
            s += 1
            e -= 1
        # pity, just add s
        else:
            s += 1
 
 
    print(sumA + bonus)

summary

This kind of pairing problem can consider sorting + greedy + Double pointer