題意：有兩個無刻度的容量分別為A,B升的杯子，通過一些操作使某一個杯子中有C升的水。

1. FILL(i) ，將i這個杯子中的水接滿
2. DROP(i)，將i這個杯子中的水倒掉
3. POUR(i,j)，將i這個杯子中的水倒入j這個杯子，能倒完就倒完，倒不完就留在杯子中。

問達到目標狀態的操作次數最少的方案是什麼？

思路：BFS+路徑輸出。共六種操作FILL(1)，FILL(2)，DROP(1)，DROP(2)，POUR(1,2)，POUR(2,1)。路徑另外去開一個數組去存來的方向就好了。如3 4這狀態是由0 4來的，path[3][4]裡面存0 4。

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 105;
int A, B, C;
struct state//狀態
{
    int a, b, step;
}NOW, NEXT;

struct PATH//路徑
{
    int a, b, op;
}path[MAXN][MAXN];

void putout(int x,int y)//路徑輸出
{
    if (x == 0 && y == 0) return ;
    putout(path[x][y].a, path[x][y].b);
    if (path[x][y].op == 1) printf("FILL(1)\n");
    else if (path[x][y].op == 2) printf("FILL(2)\n");
    else if (path[x][y].op == 3) printf("DROP(1)\n");
    else if (path[x][y].op == 4) printf("DROP(2)\n");
    else if (path[x][y].op == 5) printf("POUR(1,2)\n");
    else if (path[x][y].op == 6) printf("POUR(2,1)\n");
}

bool vis[MAXN][MAXN];
bool bfs()
{
    memset(vis, 0, sizeof(vis));
    NOW.a = 0, NOW.b = 0, NOW.step = 0; vis[0][0] = true;
    queue<state> q;
    q.push(NOW);
    while (!q.empty())
    {
        NOW = q.front();
        q.pop();
        if (NOW.a == C || NOW.b == C)//輸出答案
        {
            printf("%d\n",NOW.step);
            putout(NOW.a, NOW.b);
            return true;
        }
//FILL(i)

        //FILL(1)
        NEXT.a = A; NEXT.b = NOW.b; NEXT.step = NOW.step + 1;
        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 1;//記錄操作
        }

        //FILL(2)
        NEXT.a = NOW.a; NEXT.b = B; NEXT.step = NOW.step + 1;
        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 2;//記錄操作
        }

//DROP(i)
        //DROP(1)
        NEXT.a = 0; NEXT.b = NOW.b; NEXT.step = NOW.step + 1;
        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 3;//記錄操作
        }

        //DROP(2)
        NEXT.a = NOW.a; NEXT.b = 0; NEXT.step = NOW.step + 1;
        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 4;//記錄操作
        }

//POUR(i,j)
        //POUR(1,2)
        if (NOW.a >= B - NOW.b)//i能把j倒滿
        {
            NEXT.a = NOW.a - (B - NOW.b); NEXT.b = B; NEXT.step = NOW.step + 1;
        }
        else //倒不滿j
        {
            NEXT.a = 0; NEXT.b = NOW.b + NOW.a; NEXT.step = NOW.step + 1;
        }

        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 5;//記錄操作
        }

        //POUR(2,1)
        if (NOW.b >= A - NOW.a)//j能把i倒滿
        {
            NEXT.a = A; NEXT.b = NOW.b - (A - NOW.a); NEXT.step = NOW.step + 1;
        }
        else //倒不滿i
        {
            NEXT.a = NOW.a + NOW.b; NEXT.b = 0; NEXT.step = NOW.step + 1;
        }

        if (!vis[NEXT.a][NEXT.b])
        {
            vis[NEXT.a][NEXT.b] = true;
            q.push(NEXT);
            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//記錄路徑
            path[NEXT.a][NEXT.b].op = 6;//記錄操作
        }
    }
    return false;
}

int main()
{
    while (~scanf("%d%d%d", &A, &B, &C))
    {
        if (bfs() == false) printf("impossible\n");
    }
    return 0;
}
/*
3 4 5
*/
/*
第一個樣例解釋:
3 5 4         i  j
FILL(2)       0  5
POUR(2,1)     3  2
DROP(1)       0  2
POUR(2,1)     2  0
FILL(2)       2  5
POUR(2,1)     3  4
*/