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

 

代码：

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 using namespace std; 6 int x,y,z; 7 int cnt=0; 8 int v[200100]; 9 int ans[30];10   11 void dfs(int a, int b ,int c){12       13     if(v[a*10000+b*1000+c]==1)14     {15         return;16     }17     v[a*10000+b*1000+c]=1;18     if(a==0){19         cnt++;20         ans[c]=1;21     }22     // c--->a,b23     if(c!=0){24         //c->a25         if(a+c>=x&&a!=x) dfs(x,b,c-x+a);26         else if(a+c
         
          b28         if(b+c>=y&&b!=y) dfs(a,y,c-y+b);29         else if(b+c
          
           a,c32 if(b!=0){33 //b->a34 if(b+a>=x&&a!=x) dfs(x,b+a-x,c);35 else if(b+a
           
            c37 if(b+c>=z&&c!=z) dfs(a,b+c-z,z);38 else if(b+c
            
             c,b41 if(a!=0){42 //a->c43 if(a+c>=z&&c!=z) dfs(a+c-z,b,z);44 else if(a+c
             
              b46 if(b+a>=y&&b!=y) dfs(a+b-y,y,c);47 else if(a+b
              
               >x>>y>>z;56 memset(v,0,sizeof(v));57 memset(ans,0,sizeof(ans));58 cnt=0;59 dfs(0,0,z);60 int k=0;61 for(int i=0;i<=20;i++){62 if(ans[i]){63 if(k) cout<<" ";64 k++;65 cout<