UVa Problem Solution: 10090 - Marbles

 

This problem requires us to find two non-negative numbers m1 and m2 that satisfying

  n1*m1 + n2*m2 = n

and minimize the cost

  c = c1*m1 + c2*m2.

 

First, we can use extended Euclid's algorithm to find m1' and m2' that give

  n1*m1' + n2*m2' = g

where g is the greatest common divisor of n1 and n2.

 

Now, if g can not divide n, then there's no valid solution. Otherwise, we can get m1'' and m2'' that give

  n1*m1'' + n2*m2'' = n

by multiplying m1' and m2' by n/g.

 

Then we can write m1 and m2 in the following form

  m1 = m1'' + n2/g * t,
  m2 = m2'' - n1/g * t,
where t is a interger parameter, to investigate for which values of t the solution is feasible for our problem. Given

  m1'' + n2/g * t >= 0, m2'' - n1/g * t >= 0, n1 > 0, n2 > 0,

we can get

  -m1'' * g/n2 <= t <= m2'' * g/n1,

and since t must be an integer, we futher have

  ceil(-m1'' * g/n2) <= t <= floor(m2'' * g/n1).
So, all the feasible values of t form an interval of consecutive integer values.

Next, let's look at the cost function

  c = c1*m1 + c2*m2 = (c1*m1'' + c2*m2'') + (c1 * n2/g - c2 * n1/g) * t.

Since both (c1*m1'' + c2*m2'')  and  (c1 * n2/g - c2 * n1/g) are constant values, the cost function is in fact a linear function of t. Thus, its minimum value can occur only at the boundary of the feasible region.

So, we can just check the two boundaries t = ceil(-m1'' * g/n2) and t = floor(m2'' * g/n1), and choose the one yeilding the cheaper cost.

 

 

Code:

/*************************************************************************** * Copyright (C) 2008 by Liu Kaipeng * * LiuKaipeng at gmail dot com * ***************************************************************************//* @JUDGE_ID 00000 10090 C++ "Marbles" */#include <algorithm>#include <cstdio>#include <cstring>#include <deque>#include <fstream>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>using namespace std; typedef long long llong; llong gcd(llong a, llong b, llong& x, llong& y){ if (a < b) return gcd(b, a, y, x); if (b == 0) { x = 1; y = 0; return a; } else { llong x1, y1; llong g = gcd(b, a % b, x1, y1); x = y1; y = (x1 - a / b * y1); return g; }}bool fit(llong n, llong c1, llong n1, llong c2, llong n2, llong& m1, llong& m2){ llong g = gcd(n1, n2, m1, m2); if (n % g != 0) return false; n1 /= g, n2 /= g, n /= g; m1 *= n, m2 *= n; llong l = m1 >= 0 ? -m1 / n2 : (-m1 + n2 - 1) / n2; llong h = m2 >= 0 ? m2 / n1 : (m2 - n1 + 1) / n1; if (l > h) return false; llong cf = c1 * n2 - c2 * n1; if (cf * l <= cf *h) m1 = m1 + n2 * l, m2 = m2 - n1 * l; else m1 = m1 + n2 * h, m2 = m2 - n1 * h; return true; } int main(int argc, char *argv[]){#ifndef ONLINE_JUDGE freopen((string(argv[0]) + ".in").c_str(), "r", stdin); freopen((string(argv[0]) + ".out").c_str(), "w", stdout);#endif for (llong n, c1, n1, c2, n2, m1, m2; scanf("%lld/n", &n) && n != 0; ) { scanf("%lld %lld/n%lld %lld/n", &c1, &n1, &c2, &n2); if (fit(n, c1, n1, c2, n2, m1, m2)) printf("%lld %lld/n", m1, m2); else printf("failed/n"); } return 0;}