Looking for a faster solution to 'maximum number of teams' problem

Question

My solution exceeds the time limit and I can't come up with a faster solution, still very much a beginner. How can I improve it?

The problem:

A perfect ICPC team is made up of at least 1 mathematician and 1 programmer and it must have 3 members. You have the number of mathematicians, programmers and students that have no specialization. What is the maximum number of perfect teams you can make? C students are programmers, M students are mathematicians and X don't have a specialization.

Example input:

1 1 1

Example output:

1

Example input:

3 6 0

Example output:

3

Example input:

10 1 10

Example output:

1

My solution:

    cmx = [int(x) for x in input().split()]
    i = 0
    
    while 0 not in cmx:
        cmx[0] -= 1
        cmx[1] -= 1
        cmx[2] -= 1
        i += 1
    
    if cmx[0] != 0 and cmx[1] != 0 and cmx[2] == 0:
        while sum(cmx) >= 3 and cmx[0] != 0 and cmx[1] != 0:
            if cmx[0] >= cmx[1]:
                cmx[0] -= 2
                cmx[1] -= 1
                i += 1
            elif cmx[0] < cmx[1]:
                cmx[0] -= 1
                cmx[1] -= 2
                i += 1
    
    print(i)

Answer 1

Assume that M ? C. (Proof works identically if C ? M). How many teams can I make. It's clear that if M + C + X ? 3M, then I can easily make M teams. (Every team has a mathematician, a programmer, and either a second programmer or a "none".) And I can't make more than M teams. If M + C + X < 3M, then the most I can have is (M + C + X) / 3 teams, and again you make them the same way, since you have sufficient mathematicians and programmers.

The proof works identically if C ? M.

So min(M, C, (M + C + X) // 3). As stated above.

---

A simpler way of looking at it is that C, M, and (C + M + X)//3 are each, independently, an upper bound on the number of teams that you can form. You just have to show the smallest of these three upper bounds is, in fact, a reachable value.