(Analysis by Dhruv Rohatgi )

We can simulate the pours one by one, keeping track of the amount of milk in each bucket. For example, if we pour from bucket $1$ into bucket $2$, and bucket $1$ has size $c_1$ and bucket $2$ has size $c_2$, and before the pour, bucket $1$ has $m_1$ units of milk and bucket $2$ has $m_2$ units, then the amount of milk poured is $\min(m_1, c_2 - m_2)$. Therefore after the pour, the amount of milk in bucket $1$ is $m_1 - \min(m_1, c_2 - m_2)$. And the amount of milk in bucket $2$ is $m_2 + \min(m_1, c_2 - m_2)$. The formulas for pouring bucket $2$ into bucket $3$, or bucket $3$ into bucket $1$, are analogous.

Since there are only $100$ pours, and each pour takes only a constant number of arithmetic operations to simulate, this algorithm will run very fast.

Here is Travis Hance's code:

#include <cstdio>
#include <algorithm>
using namespace std;

void pour(int& c1, int& m1, int& c2, int& m2) {
int amt = min(m1, c2 - m2);
m1 -= amt;
m2 += amt;
}

int main() {
int c1, c2, c3;
int m1, m2, m3;
scanf("%d %d", &c1, &m1);
scanf("%d %d", &c2, &m2);
scanf("%d %d", &c3, &m3);

for (int i = 0; i < 33; i++) {
pour(c1, m1, c2, m2);
pour(c2, m2, c3, m3);
pour(c3, m3, c1, m1);
}
pour(c1, m1, c2, m2);

printf("%d\n%d\n%d\n", m1, m2, m3);
}