1 /*题意:要求将一瓶可乐平均分成份，问能否平均分成两份，如果能输出最少需要几次否则输出NO  2 题目给出三个整数 S N M  S表示可乐总量   N M分别为两个杯子的容量 且 S= N + M  3   4 分析:题目是由 S 0 0 状态 转换到 0 S/2 S/2 的状态且要找出最优解，用广搜  5      题目只有6种状态转换方式，建立一格六叉数的模型进行遍历。  6      用结构体记录步数 步数为弹出队头元素的步数+1  7      遍历完成条件为 1.找到一个方法能够平分 2.遍历所有可能情况仍然无法找到  8   9 注意: 分清楚 S N M 在标记中的位置 10 */ 11 #include 
      
        12 #include 
       
         13 using namespace std; 14 struct FUN 15 { 16     int s,n,m; 17     int step; 18 }; 19 int s,n,m; 20 int used[110][110][110];//used[s][n][m] 21 int BFS() 22 { 23     memset(used,0,sizeof(used)); 24     queue 
        
          q; 25     FUN a; 26     a.s = s; 27     a.n = 0; 28     a.m = 0; 29     a.step = 0; 30     used[a.s][a.n][a.m]=1; 31     q.push(a); 32     while(!q.empty()) 33     { 34         FUN b; 35         a = q.front(); 36         a.step = a.step + 1; 37         q.pop(); 38         //s -> n 39         b = a; 40         if(b.s > n - b.n)     //能装满 41         { 42             b.s = b.s - (n - b.n); 43             b.n = n; 44         } 45         else                 //不能装满 46         { 47             b.n = b.n + b.s; 48             b.s = 0; 49         } 50         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0) 51         { 52             return b.step; 53         } 54         if(!used[b.s][b.n][b.m]) 55         { 56             used[b.s][b.n][b.m] = 1; 57             q.push(b); 58         } 59  60         //s -> m 61         b = a; 62         if(b.s > m - b.m) 63         { 64             b.s = b.s - (m - b.m); 65             b.m = m; 66         } 67         else 68         { 69             b.m = b.m + b.s; 70             b.s = 0; 71         } 72         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0) 73         { 74             return b.step; 75         } 76         if(!used[b.s][b.n][b.m]) 77         { 78             used[b.s][b.n][b.m] = 1; 79             q.push(b); 80         } 81  82         //n -> m 83         b = a; 84         if(b.n > m - b.m) 85         { 86             b.n = b.n - (m - b.m); 87             b.m = m; 88         } 89         else 90         { 91             b.m = b.m + b.n; 92             b.n = 0; 93         } 94         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0) 95         { 96             return b.step; 97         } 98         if(!used[b.s][b.n][b.m]) 99         {100             used[b.s][b.n][b.m] = 1;101             q.push(b);102         }103 104         //n -> s105         b = a;106         if(b.n > s - b.s)107         {108             b.n = b.n - (s - b.s);109             b.s = s;110         }111         else112         {113             b.s = b.s + b.n;114             b.n = 0;115         }116         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0)117         {118             return b.step;119         }120         if(!used[b.s][b.n][b.m])121         {122             used[b.s][b.n][b.m] = 1;123             q.push(b);124         }125 126         //m -> s127         b = a;128         if(b.m > s - b.s)129         {130             b.m = b.m - (s - b.s);131             b.s = s;132         }133         else134         {135             b.s = b.s + b.m;136             b.m = 0;137         }138         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0)139         {140             return b.step;141         }142         if(!used[b.s][b.n][b.m])143         {144             used[b.s][b.n][b.m] = 1;145             q.push(b);146         }147 148         //m -> n149         b = a;150         if(b.m > n - b.n)151         {152             b.m = b.m - (n - b.n);153             b.n = n;154         }155         else156         {157             b.n = b.n + b.m;158             b.m = 0;159         }160         if(b.m == b.n && b.s == 0 || b.m == b.s &&  b.n == 0 || b.n == b.s &&  b.m == 0)161         {162             return b.step;163         }164         if(!used[b.s][b.n][b.m])165         {166             used[b.s][b.n][b.m] = 1;167             q.push(b);168         }169     }170     return 0;171 }172 int main()173 {174     while(cin>>s>>n>>m,s+n+m)175     {176         if(s % 2)177         {178             cout<<"NO"<