Codechef2015 May

用后缀自动机统计出出现1~n次的串的数量f[i]

对于ans[k]=sigma（f[i]*C(i,k)) i>=k

const maxn=10008;
      mo=1000000007;
var    a,f,rig:array[0..maxn] of dword;
    nt:array[0..maxn,'a'..'z'] of longint;
    last,sum,i:dword;
    s:ansistring;
    eg:array[0..maxn*2] of record nt,v:dword; end;
    lt:array[0..maxn] of dword;
    el:dword;
    
    time:array[0..maxn] of qword;
    T,j,TT:dword;
    ans:array[0..maxn] of qword;
    C:array[-1..5000,-1..5000] of qword;
    du,g:array[0..maxn] of longint;
    b:array[0..1000000] of longint;
procedure SAM_init; inline;
begin
    fillchar(f,sizeof(f),255);
    fillchar(nt,sizeof(nt),255);
    fillchar(a,sizeof(a),0);
    fillchar(rig,sizeof(rig),0);
    el:=0;
    fillchar(lt,sizeof(lt),0);
    fillchar(ans,sizeof(ans),0);
    fillchar(du,sizeof(du),0);
    fillchar(g,sizeof(g),0);
    last:=0; sum:=0;
end;

procedure SAM_ins(ch:char); inline;
var next,p,np,q,nq:longint;
begin
    inc(sum); p:=last; np:=sum; a[np]:=a[p]+1; last:=np;  rig[np]:=1;
    while (p<>-1) and (nt[p][ch]=-1) do begin nt[p][ch]:=np; p:=f[p]; end;
    if p=-1 then f[np]:=0 else
    begin
        q:=nt[p][ch];
        if a[p]+1=a[q] then f[np]:=q else
        begin
            inc(sum); nq:=sum; a[nq]:=a[p]+1;
            nt[nq]:=nt[q];
            f[nq]:=f[q]; f[q]:=nq; f[np]:=nq;
            while (p<>-1) and (nt[p][ch]=q) do begin nt[p][ch]:=nq; p:=f[p]; end;
        end;
    end;
end;

procedure SAM_visit1; inline;
var i,l,r:longint;
    c:char;
begin
    for i:=0 to sum do
        for c:='a' to 'z' do
            if nt[i][c]<>-1 then inc(du[nt[i][c]]);
    
    
    l:=1; r:=1; b[1]:=0; g[0]:=1;
    while l<=r do
    begin
        for c:='a' to 'z' do
            if nt[b[l]][c]<>-1 then
            begin
                dec(du[nt[b[l]][c]]);
                inc(g[nt[b[l]][c]],g[b[l]]);
                if du[nt[b[l]][c]]=0 then
                begin
                    inc(r);
                    b[r]:=nt[b[l]][c];
                end;
            end;
        inc(l);
    end;
    for i:=1 to sum do inc(time[rig[i]],g[i]);
end;
procedure dfs(u:dword);
var i:dword;
begin
    i:=lt[u];
    while i<>0 do
    begin
        dfs(eg[i].v);
        rig[u]:=rig[u]+rig[eg[i].v];
        i:=eg[i].nt;
    end;
end;
procedure add(u,v:dword); inline;
begin
    inc(el);
    eg[el].v:=v;
    eg[el].nt:=lt[u];
    lt[u]:=el;
end;
procedure SAM_rig; inline;
begin
    el:=0;
    fillchar(lt,sizeof(lt),0);
    for i:=1 to sum do add(f[i],i);
    dfs(0);
end;
procedure main; inline;
var n,q,i,j:dword;
    cnt:qword;
begin
    if T<>1 then
    begin    
        fillchar(time,sizeof(time),0);
        SAM_init;
    end;
    readln(n,q);
    readln(s);
    SAM_init;
    for i:=1 to n do SAM_ins(s[i]);
    SAM_rig;
    SAM_visit1;
    for i:=1 to n do
        for j:=i to n do ans[i]:=(ans[i]+C[j,i]*time[j]) mod mo;
    for i:=1 to q do
    begin
        readln(j);
        if j>n then writeln(0) else
        writeln(ans[j]);
    end;
end;
begin
    C[0,0]:=1;
    for i:=1 to 5000 do
        for j:=0 to i do
        begin
            C[i,j]:=(c[i-1,j-1]+c[i-1,j]);
            if C[i,j]>=mo then C[i,j]:=C[i,j]-mo;
        end;
    readln(T);
    for TT:=1 to T do main;
end.