记录编号	39425	评测结果	AAAAAAAAAA
题目名称	[HNOI 1999] 快餐问题	最终得分	100
用户昵称	czp	是否通过	通过
代码语言	Pascal	运行时间	0.837 s
提交时间	2012-07-10 18:06:02	内存使用	0.60 MiB

var
 a,b,c,i,j,n,p1,p2,p3,ans,d,l,k1,k2,k3,k,o1,o2,pr1,pr2,pp1,pp2,bo,bi,la,lb,lc,up,sum:longint;
 pv:boolean;
 v:array[1..10] of longint;
 f:array[0..10,0..101,0..101] of longint;
function big(xx,yy:longint):longint;
begin if xx>yy then big:=xx else big:=yy; end;
procedure make(k1,k2,k3:longint);
var o1:longint;
begin
      o1:=k1 div a;
      if k2 div b<o1 then o1:=k2 div b;
      if k3 div c<o1 then o1:=k3 div c;
      if o1>o2 then o2:=o1;
      if o2=up then pv:=true;
end;
begin
 assign(input,'meal.in');reset(input);
 assign(output,'meal.out');rewrite(output);
 readln(a,b,c);
 readln(p1,p2,p3);
 la:=100;lb:=100;lc:=100;
 d:=a*p1+b*p2+c*p3;
 readln(n);
 for i:=1 to n do
  begin
  read(v[i]);
  for j:=1 to 1000 do
   if (j*d>v[i])or (la-a*j<0)
   or(lb-b*j<0) or (lc-c*j<0)  then break;
  dec(j);
  dec(v[i],j*d);
  if (la-a*j>=0) and(lb-b*j>=0) and(lc-c*j>=0 ) then
  begin
  la:=la-a*j;
  lb:=lb-b*j;
  lc:=lc-c*j;
  inc(ans,j);
  end;
  inc(sum,v[i]);
  end;
 up:=sum div d;
 if up>100 div a then up:=100 div a;
 if up>100 div b then up:=100 div b;
 if up>100 div c then up:=100 div c;
 readln;
 fillchar(f,sizeof(f),$ff);
 f[0,0,0]:=0;
 pr1:=0;pr2:=0; pv:=false;
 for l:=1 to n do
  begin
   for i:=0 to 100 do
    begin
     if pv then break;
     if i>la-(la mod a)+1 then break;
     if p1*i>v[l] then break;
      for j:=0 to 100 do
        begin
         if pv then break;
         if j>lb-(lb mod b)+1 then break;
         if p1*i+p2*j>v[l] then break;
         k:=(v[l]-p1*i-p2*j) div p3;
         for k1:=0 to 100 do
          begin
          if k1>pr1 then break;
          for k2:=0 to 100 do
            begin
              if k2>pr2 then break;
              if f[l-1,k1,k2]>-1 then
                 begin
                  f[l,k1,k2]:=big(f[l,k1,k2],f[l-1,k1,k2]);
                  if (k1+i<=la)and(k2+j<=lb)and(k3+k<=lc)then  begin
                  f[l,k1+i,k2+j]:=big(f[l,k1+i,k2+j],f[l-1,k1,k2]+k);
                  make(k1+i,k2+j,f[l,k1+i,k2+j]);
                  if k1+i>pp1 then pp1:=k1+i;
                  if k2+j>pp2 then pp2:=k2+j;  end;
                 end;
            end;
          end;
       end;
    end;
    pr1:=pp1;
    pr2:=pp2;
  end;
  writeln(ans+o2);
  close(input);close(output);
end.