描述

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.

輸入

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

輸出

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.

樣例輸入

8 9 10

2 5 10

樣例輸出

1 2 8 9 10
5 6 7 8 9 10

題目大意：有A,B,C三個杯子,A,B為空杯子，C杯子開始裝滿了牛奶。 一個杯子裏的牛奶可以往另外兩個杯子裏到(直到杯內牛奶倒完或者另一個杯子牛奶裝滿)

求A杯子空的時候C杯子牛奶可能的情況。 解題思路：深搜

#include <bits/stdc++.h>
using namespace std;
int a[65][65][65],b[65];
int n,m,k;
void dfs(int A,int B,int C)
{
    if(a[A][B][C]) return
 
;
    a[A][B][C]=1;
    if(A==0) b[C]=1;
    if(A>0)
    {
        int mm=min(m-B,A);
        dfs(A-mm,B+mm,C);
        mm=min(k-C,A);
        dfs(A-mm,B,C+mm);
    }
    if(B>0)
    {
        int mm=min(B,n-A);
        dfs(A+mm,B-mm,C);
        mm=min(B,k-C);
        dfs(A,B-mm,C+mm);
    }
    if(C>0)
    {
        int mm=min(C,n-A);
        dfs(A+mm,B,C-mm);
        mm=min(C,m-B);
        dfs(A,B+mm,C-mm);
    }
}
int main()
{
    while(cin>>n>>m>>k)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        dfs(0,0,k);
        int f=1;
        for(int i=0;i<=60;i++)
        {
            if(b[i]&&f)
            {
                cout<<i;
                f=0;
            }
            else if(b[i]) cout<<" "<<i;
        }
        cout<<"\n";
    }
}

Mother's Milk