Print three lines of output, giving the final amount of milk in each bucket, after 100 pour operations.
My 1st Correct Code
#include <iostream>
int main() {
int c[4]; int m[4];
std::cin >> c[1] >> m[1];
std::cin >> c[2] >> m[2];
std::cin >> c[3] >> m[3];
for (int i = 0; i < 100; i++) {
int from = i % 3 + 1;
int to = (i + 1) % 3 + 1;
int tmp;
tmp = c[to] - m[to];
if (m[from] >= tmp) {
m[to] += tmp; m[from] -= tmp;
}
else {
m[to] += m[from]; m[from] = 0;
}
}
printf("%d\n%d\n%d", m[1], m[2], m[3]);
}
comments
2020-01-16 …
