应该比较好看出来是费用流，那么就考虑怎么构图

首先我们把一天拆成两个点，XI,YI，分别代表这一天买了多少

和洗多少，再加入源和汇S,T

1.每一天我们可以买新的毛巾，所以连接一条从S到XI的边，流量为正无穷（因为可以买好多），费用为f

2.然后我们对于买来的毛巾可以洗，每天都产生need[i]的毛巾可以洗，那么连一条从S到YI的边，

流量为need[i]，费用为0（因为只决定要洗，没有确定洗的方案，所以先不算费用）

3.每一条要用a方法洗的毛巾，我们连一条从YI到X(I+a+1)的边，流量为正无穷（下文解释），费用为fa的

4.每一条要用b方法洗的毛巾，我们连一条从YI到X(I+b+1)的边，流量为正无穷（下文解释），费用为fb的

5.因为每天剩下的毛巾，我们可以不当天洗，所以连接一条从YI到Y(I+1)的边，流量为正无穷，费用为0(所以每天可以洗

的毛巾的个数是可能会很多的，4,5建的边流量要是正无穷)

6.那么我们每天买的毛巾除了洗，还可以不洗，也就是直接扔掉，所以连一条XI到T的边，流量为need[i]，费用为0

/**************************************************************    Problem: 1221    User: BLADEVIL    Language: Pascal    Result: Accepted    Time:1952 ms    Memory:8120 kb****************************************************************/ //By BLADEVILvar    n, a, b, f, fa, fb  :longint;    need                :array[0..1010] of longint;    pre, other,len      :array[0..500010] of longint;    cost                :array[0..500010] of longint;    last                :array[0..4010] of longint;    ss, tt              :longint;    que                 :array[0..5000] of longint;    dis                 :array[0..5000] of longint;    flag                :array[0..5000] of boolean;    father              :array[0..5000] of longint;    ans                 :int64;    l                   :longint;      function min(a,b:longint):longint;begin    if a>b then min:=b else min:=a;end;      procedure connect(a,b,c,d:longint);begin    inc(l);    pre[l]:=last[a];    last[a]:=l;    other[l]:=b;    len[l]:=c;    cost[l]:=d;end;      procedure init;var    i                   :longint;    use                 :longint;begin    read(n,a,b,f,fa,fb);    for i:=1 to n do read(need[i]);    l:=1;    ss:=2*n+1;    tt:=2*n+2;    for i:=1 to n do    begin        connect(ss,2*i-1,maxlongint,f);        connect(2*i-1,ss,0,-f);        connect(2*i-1,tt,need[i],0);        connect(tt,2*i-1,0,0);        connect(ss,2*i,need[i],0);        connect(2*i,ss,0,0);        use:=i+a+1;        if use<=n then        begin            use:=use*2-1;            connect(2*i,use,maxlongint,fa);            connect(use,2*i,0,-fa);        end;        use:=i+b+1;        if use<=n then        begin            use:=use*2-1;            connect(2*i,use,maxlongint,fb);            connect(use,2*i,0,-fb);        end;    end;    for i:=1 to n-1 do    begin        connect(2*i,2*i+2,maxlongint,0);        connect(2*i+2,2*i,0,0);    end;end;  function spfa:boolean;var    q, p, cur               :longint;    h, t                    :longint;begin    filldword(dis,sizeof(dis) div 4,maxlongint div 10);    que[1]:=ss;    dis[ss]:=0;    h:=0; t:=1;    while t<>h do    begin        h:=h mod 5000+1;        cur:=que[h];        flag[cur]:=false;        q:=last[cur];        while q<>0 do        begin            p:=other[q];            if len[q]>0 then            begin                if dis[p]>dis[cur]+cost[q] then                begin                    father[p]:=q;                    dis[p]:=dis[cur]+cost[q];                    if not flag[p] then                    begin                        t:=t mod 5000+1;                        que[t]:=p;                        flag[p]:=true;                    end;                end;            end;            q:=pre[q];        end;    end;    if dis[tt]=maxlongint div 10 then exit(false) else exit(true);end;  procedure update;var    cur                     :longint;    low                     :longint;   begin    low:=maxlongint div 10;    cur:=tt;    while cur<>ss do    begin        low:=min(low,len[father[cur]]);        cur:=other[father[cur] xor 1];    end;    cur:=tt;    while cur<>ss do    begin        dec(len[father[cur]],low);        inc(len[father[cur] xor 1],low);        inc(ans,low*cost[father[cur]]);        cur:=other[father[cur] xor 1];     end;end;  procedure main;begin    while spfa do update;    writeln(ans);end;  begin    init;    main;end.