1 #include <stdio.h>
  2 #include <string.h>
  3 #include <queue>
  4 #include <algorithm>
  5 using namespace std;
  6 struct node
  7 {
  8     int S，N,M;   //三个杯子存在的可乐量      
  9     int count;
 10 };
 11 int s,n,m;          //三个杯子的容量
 12 int res = -1;       //初始化最后结果
 13 int vis[500][500];      //访问标记
 14 queue <node> q;         //定义一个队列
 15 void bfs()
 16 {
 17     memset(vis,0,sizeof(vis));
 18     node now;
 19     q.push((node){s,0,0,0});
 20     vis[0][0] = 1;
 21     while (!q.empty())
 22     {
 23         now = q.front();
 24         //printf("%d %d %d %d\n",now.S,now.N,now.M,now.count);
 25         if ( (now.S == s/2 && now.N == s/2) || (now.N == s/2 && now.M == s/2) || (now.M == s/2 && now.S == s/2) )           //如果已经评分 把值赋予给res
 26         {
 27             res = now.count;
 28             while (!q.empty())          //清空队列
 29             {
 30                 q.pop();
 31             }
 32             return ;
 33         }
 34         q.pop();
 35         if (now.N < n)              //如果第二杯子没有满
 36         {
 37             if (now.S >= n - now.N && !vis[n][now.M])           //如果第一个杯子可以装满第二个杯子
 38             {
 39                 vis[n][now.M] = 1;
 40                 q.push((node){now.S - n + now.N, n , now.M , now.count+1});
 41                 
 42             }
 43             if ( now.S < n-now.N && !vis[now.N+now.S][now.M])       //如果第一个杯子中的量装不满第二个
 44             {
 45                 vis[now.N+now.S][now.M] = 1;
 46                 q.push((node){0,now.N + now.S,now.M , now.count+1});
 47             }
 48             if (now.M >= n - now.N && !vis[n][now.M - n + now.N])       //第三个杯子中的量可以装满第二个杯子
 49             {
 50                 vis[n][now.M - n + now.N] = 1;
 51                 q.push((node){now.S,n,now.M - n + now.N,now.count+1});
 52                 
 53             }
 54             if ( now.M < n-now.N && !vis[now.N+now.M][0])           //第三个杯子中的量不可以装满第二个杯子
 55             {
 56                 vis[now.N+now.M][0] = 1;
 57                 q.push((node){now.S ,  now.N + now.M , 0 ,now.count+1});
 58             }
 59         }
 60         if (now.M < m)                                              //后面的情况类似
 61         {
 62             if (now.S >= m - now.M && !vis[now.N][m])
 63             {
 64                 vis[now.N][m] = 1;
 65                 q.push((node){now.S - m + now.M,  now.N  , m ,now.count+1});
 66             }
 67             if ( now.S < m-now.M && !vis[now.N][now.M + now.S])
 68             {
 69                 vis[now.N][now.M + now.S] = 1;
 70                 q.push((node){0,  now.N  , now.M + now.S ,now.count+1});
 71                 
 72             }
 73             if (now.N >= m - now.M && !vis[now.N - m + now.M][m])
 74             {
 75                 vis[now.N - m + now.M][m] = 1;
 76                 q.push((node){now.S ,  now.N - m + now.M  , m ,now.count+1});
 77                 
 78             }
 79             if ( now.N < m-now.M && !vis[0][now.M + now.N])
 80             {
 81                 vis[0][now.M + now.N] = 1;
 82                 q.push((node){now.S ,  0  , now.N + now.M ,now.count+1});
 83                 
 84             }
 85         }
 86         if (now.S < s)                          //因为第一个杯子最大，，所以不存在后面的某个杯子能装满 所以不用讨论
 87         {
 88             if ( now.N < s - now.S && !vis[0][now.M])
 89             {
 90                 vis[0][now.M] = 1;
 91                 q.push((node){now.S + now.N ,  0  , now.M ,now.count+1});
 92                 
 93             }
 94             if ( now.M < s-now.S && !vis[now.N][0])
 95             {
 96                 vis[now.N][0] = 1;
 97                 q.push((node){now.S + now.M ,  now.N  , 0 ,now.count+1});
 98                 
 99             }
100         }
101     }
102 
103 }
104 int main()
105 {
106     while (~scanf("%d%d%d",&s, &n, &m ) )
107     {
108         res = -1;
109         if ( s == 0 && n == 0 && m == 0)
110         {
111             break;
112         }
113         if (s % 2 != 0 )            //如果总量是奇数，不可能被平分
114         {
115             printf("NO\n");
116             continue;
117         }
118         bfs();
119         if ( res == -1 )
120         {
121             printf("NO\n");
122             continue;
123         }
124         else 
125         {
126             printf("%d\n",res);
127         }
128     }
129     return 0;
130 }