USACO 2018 December Contest Bronze Division - Mixing Milk

Problem link: here

Solution Author: Stefan Dascalescu

Problem Solution

We have at most 100 moves, so we can simulate the process and keep track of how much milk we have by the time the process ends.

Source codes

The source codes in C++ and Python can be seen below.

#include <bits/stdc++.h>
using namespace std;
ifstream f("mixmilk.in");
ofstream g("mixmilk.out");
int v[5], cap[5];
int main()
{
    for(int i = 1; i <= 3; ++i)
        f >> cap[i] >> v[i];
    for(int i = 1; i <= 100; ++i)
    {
        if(i % 3 != 0)
        {
            int x = (i % 3);
            int aa = min(v[x], cap[x+1] - v[x+1]);
            v[x] -= aa;
            v[x+1] += aa;
        }
        else
        {
            int aa = min(v[3], cap[1] - v[1]);
            v[3] -= aa;
            v[1] += aa;
        }
    }
    for(int i = 1; i <= 3; ++i)
        g << v[i] << '\n';
    return 0;
}

with open("mixmilk.in") as fin:
    tokens = fin.read().split()

cap = [0] * 4
v = [0] * 4
for i in range(1, 4):
    cap[i] = int(tokens[2*(i-1)])
    v[i] = int(tokens[2*(i-1)+1])

for i in range(1, 101):
    if i % 3 != 0:
        x = i % 3
        aa = min(v[x], cap[x+1] - v[x+1])
        v[x] -= aa
        v[x+1] += aa
    else:
        aa = min(v[3], cap[1] - v[1])
        v[3] -= aa
        v[1] += aa

with open("mixmilk.out", "w") as fout:
    for i in range(1, 4):
        fout.write(str(v[i]) + "\n")