This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

Farmer John prides himself on having the healthiest dairy cows in the world. He knows the vitamin content for one scoop of each feed type and the minimum daily vitamin requirement for his cows. Help Farmer John feed the cows so they stay healthy while minimizing the number of scoops that a cow is fed.

Given the daily requirements of each kind of vitamin that a cow needs, identify the smallest combination of scoops of feed a cow can be fed in order to meet at least the minimum vitamin requirements.

Vitamins are measured in integer units. Cows can be fed at most one scoop of any feed type. It is guaranteed that a solution exists for all contest input data.

### INPUT FORMAT

 Line 1: integer V (1 <= V <= 25), the number of types of vitamins Line 2: V integers (1 <= each one <= 1000), the minimum requirement for each of the V vitamins that a cow requires each day Line 3: integer G (1 <= G <= 15), the number of types of feeds available Lines 4..G+3: V integers (0 <= each one <= 1000), the amount of each vitamin that one scoop of this feed contains. The first line of these G lines describes feed #1; the second line describes feed #2; and so on.

### SAMPLE INPUT (file holstein.in)

``````4
100 200 300 400
3
50   50  50  50
200 300 200 300
900 150 389 399
``````

### OUTPUT FORMAT

The output is a single line of output that contains:

• the minimum number of scoops a cow must eat, followed by:
• a SORTED list (from smallest to largest) of the feed types the cow is given

If more than one set of feedtypes yield a minimum of scoops, choose the set with the smallest feedtype numbers.

### SAMPLE OUTPUT (file holstein.out)

``````2 1 3
``````

Java

C++

Pascal

## ANALYSIS

Hal Burch

Since there are only 15 feeds, and for each feed we can either give zero or one scopes of it, there are 215 possible ‘feed mixtures’ the cows can be fed, which is only 32,768. Therefore, try all combinations and pick which of the legal combinations uses the least number of feeds.

``````#include <stdio.h>
#include <assert.h>

#define MAXV 25
#define MAXF 15

int req[MAXV]; /* the vitamin requirements */
int numv; /* number of vitamins */

int feeds[MAXF][MAXV]; /* the vitamin within each feed */
int numf; /* number of feeds */

int best; /* the minimum number of feeds to use found thus far */
int bestf[MAXF]; /* the set */

int curf[MAXF]; /* the current set of feeds being considered */

void find_feed(int fcnt, int fid)
{ /* fcnt is the number of feeds in the current mixture,
fid is the identifier of the first feed to try adding (last feed + 1) */
int lv;

/* check if the requirement has been met */
for (lv = 0; lv < numv; lv++)
if (req[lv] > 0) break;
if (lv >= numv)
{ /* all the requirements are met */
/* we know this is better, since we wouldn't have checked it otherwise
(see below) */
best = fcnt;
for (lv = 0; lv < best; lv++)
bestf[lv] = curf[lv];
return;
}

while (fid < numf && fcnt+1 < best)
{ /* try adding each feed to the mixture */
/* the fcnt+1 < best ensures that we stop if there's no hope
in finding a better solution than one found already */

/* add the vitamins from this feed */
for (lv = 0; lv < numv; lv++)
req[lv] -= feeds[fid][lv];
curf[fcnt] = fid; /* put it in the list */

find_feed(fcnt+1, fid+1);

/* undo adding the vitamins */
for (lv = 0; lv < numv; lv++)
req[lv] += feeds[fid][lv];

/* next feed */
fid++;
}
}

int main(void)
{
FILE *fin, *fout;
int lv, lv2;

fin = fopen("holstein.in", "r");
fout = fopen("holstein.out", "w");
assert(fin);
assert(fout);

fscanf (fin, "%d", &numv);
for (lv = 0; lv < numv; lv++)
fscanf (fin, "%d", &req[lv]);
fscanf (fin, "%d", &numf);
for (lv = 0; lv < numf; lv++)
for (lv2 = 0; lv2 < numv; lv2++)
fscanf (fin, "%d", &feeds[lv][lv2]);

best = numf+1;
find_feed(0, 0);

fprintf (fout, "%i", best);
for (lv = 0; lv < best; lv++)
fprintf (fout, " %i", bestf[lv]+1);
fprintf (fout, "\n");
return 0;
}
``````