记录一个菜逼的成长。。
就是把操作用string保存。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <deque>
#include <cctype>
#include <bitset>
#include <cmath>
using namespace std;
#define ALL(v) (v).begin(),(v).end()
#define cl(a,b) memset(a,b,sizeof(a))
#define bp __builtin_popcount
#define pb push_back
#define mp make_pair
#define fin freopen("D://in.txt","r",stdin)
#define fout freopen("D://out.txt","w",stdout)
#define lson t<<1,l,mid
#define rson t<<1|1,mid+1,r
#define seglen (node[t].r-node[t].l+1)
#define pi 3.1415926
#define exp 2.718281828459
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef vector<PII> VPII;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename T>
inline void read(T &x){
T ans=0;
char last=' ',ch=getchar();
while(ch<'0' || ch>'9')last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans;
x = ans;
}
inline bool DBread(double &num)
{
char in;double Dec=0.1;
bool IsN=false,IsD=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))
in=getchar();
if(in=='-'){IsN=true;num=0;}
else if(in=='.'){IsD=true;num=0;}
else num=in-'0';
if(!IsD){
while(in=getchar(),in>='0'&&in<='9'){
num*=10;num+=in-'0';}
}
if(in!='.'){
if(IsN) num=-num;
return true;
}else{
while(in=getchar(),in>='0'&&in<='9'){
num+=Dec*(in-'0');Dec*=0.1;
}
}
if(IsN) num=-num;
return true;
}
template <typename T>
inline void write(T a) {
if(a < 0) { putchar('-'); a = -a; }
if(a >= 10) write(a / 10);
putchar(a % 10 + '0');
}
/******************head***********************/
int A,B,C;
struct node{
int x,y,step;
string ope;
}ans;
int vis[110][110];
string s[] = {"","FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
int bfs()
{
cl(vis,0);
queue<node>q;
node a,b;
a.x = a.y = a.step = 0;
a.ope = '0';
vis[0][0] = 1;
q.push(a);
while(!q.empty()){
node f = q.front();
q.pop();
//cout<<f.ope<<endl;
if(f.x == C || f.y == C){
ans = f;
return 1;
}
b = f;
b.x = A;
b.ope += '1';
b.step++;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
b = f;
b.y = B;
b.ope += '2';
b.step++;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
b = f;
b.x = 0;
b.ope += '3';
b.step++;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
b = f;
b.y = 0;
b.ope += '4';
b.step++;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
b = f;
b.ope += '5';
b.step++;
if(b.x + b.y >= B){
b.x -= (B - b.y);//这两步不能弄反,就是因为这两步弄反了,整个人都快疯了。。
b.y = B;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
}
else {
b.y += b.x;
b.x = 0;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
}
b = f;
b.ope += '6';
b.step++;
if(b.x+b.y >= A){
b.y -= (A - b.x);//这两步不能弄反,就是因为这两步弄反了,整个人都快疯了。。
b.x = A;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
}
else {
b.x += b.y;
b.y = 0;
if(!vis[b.x][b.y]){
vis[b.x][b.y] = 1;
q.push(b);
}
}
}
return 0;
}
int main()
{
while(~scanf("%d%d%d",&A,&B,&C)){
int ret = bfs();
if(ret){
printf("%d\n",ans.step);
for( int i = 1; i < ans.ope.size(); i++ ){
cout<<s[ans.ope[i] - '0']<<endl;
}
}
else puts("impossible");
}
return 0;
}