非常可乐

大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。

Input

三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。

Output

如果能平分的话请输出最少要倒的次数，否则输出"NO"。

Sample Input

7 4 3
4 1 3
0 0 0

Sample Output

NO
3

bfs 模拟即可，需要注意的是，只要三个容器里任意两个相等即可，不一定要杯子两个相等

#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int Max = 101;
//一共三个杯子，知道总和，所以只需要前两个杯子的状态，就可以进行判断了
int s,n,m;
int vis[Max][Max];
typedef struct node {
    int x,y,z;//x代表可乐罐，y,z代表两个杯子
    int step;
}Node;

//模拟的顺序是：x to y,x to z,y to x,y to z,z to x,z to y
int bfs() {
    memset(vis,0,sizeof(vis));
    Node now,next;
    queue<Node> q;
    now.x = s,now.y = 0,now.z = 0,now.step = 0;
    vis[now.x][now.y] = 1;
    q.push(now);
    while(q.size()) {
        now = q.front();
        q.pop();
        if((now.x == now.y && now.z == 0) || (now.x == now.z && now.y == 0) || (now.y == now.z && now.x == 0))
            return now.step;
        for(int i = 0;i < 6;i++) {
            switch(i) {
                case 0:
                    if(now.x != 0 && now.y != n)
                        next.x = now.x - (n - now.y),next.y = n,next.z = now.z;
                    break;
                case 1:
                    if(now.x != 0 && now.z != m)
                        next.x = now.x - (m - now.z),next.y = now.y,next.z = m;
                    break;
                case 2:
                    if(now.y != 0)
                        next.x = now.x + now.y,next.y = 0,next.z = now.z;
                    break;
                case 3:
                    if(now.y != 0 && now.z != m) {
                        if(now.y + now.z > m)
                            next.x = now.x,next.y = now.y - (m - now.z),next.z = m;
                        else
                            next.x = now.x,next.y = 0,next.z = now.z + now.y;
                    }
                    break;
                case 4: 
                    if(now.z != 0)
                        next.x = now.x + now.z,next.y = now.y,next.z = 0;
                    break;
                case 5: 
                        if(now.z != 0 && now.y != n) {
                            if(now.y + now.z > n)
                                next.x = now.x,next.y = n,next.z = now.z - (n - now.y);
                            else
                                next.x = now.x,next.y = now.y + now.z,next.z = 0;
                        }
                        break;
            }
            if(vis[next.x][next.y])
                continue;
            vis[next.x][next.y] = 1;
            next.step=now.step + 1;
            q.push(next);
        }
    }
    return -1;
}

int main() {
    while(~scanf("%d %d %d",&s,&n,&m)) {
        if(!s && !n && !m)
            break;
        memset(vis,0,sizeof(vis));
        int ans = bfs();
        if(ans == -1)
            printf("NO\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}