学步园

Mother's Milk

Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.

Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.

PROGRAM NAME: milk3

INPUT FORMAT

A single line with the three integers A, B, and C.

SAMPLE INPUT (file milk3.in)

8 9 10

OUTPUT FORMAT

A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.

SAMPLE OUTPUT (file milk3.out)

1 2 8 9 10

SAMPLE INPUT (file milk3.in)

2 5 10

SAMPLE OUTPUT (file milk3.out)

5 6 7 8 9 10
#include <stdio.h>

int m[3];

int s[21][21][21];

#define min(a,b) (a>b?b:a)

void dfs(int a, int b, int c)
{
    if (a<0 || b<0 || c<0) return;
    if (!s[a][b][c]) {
        s[a][b][c] = 1;
        dfs(a+min(b,m[0]-a), b-min(b,m[0]-a), c);
        dfs(a+min(c,m[0]-a), b, c-min(c,m[0]-a));

        dfs(a-min(a,m[1]-b), b+min(a,m[1]-b), c);
        dfs(a, b+min(c,m[1]-b), c-min(c,m[1]-b));

        dfs(a-min(a,m[2]-c), b, c+min(a,m[2]-c));
        dfs(a, b-min(b,m[2]-c), c+min(b,m[2]-c));
    }
}

int r[21],t;

int main()
{
    int i,j;
    FILE *f = fopen("milk3.in", "r");
    fscanf(f, "%d %d %d", &m[0], &m[1], &m[2]);
    dfs(0,0,m[2]);
    f = fopen("milk3.out", "w");
    for(j=0;j<21; ++j)
        for(i=0; i<21; ++i)
            if (s[0][i][j])
                r[t++] = j;
    for(j=0; j<t-1; ++j) fprintf(f, "%d ", r[j]);
    if (j) fprintf(f, "%d/n", r[j]);
    return 0;
}