bzoj 1089 SCOI2003严格n元树 递推

挺好想的，就是一直没调过，我也不知道哪儿的错，对拍也拍了，因为数据范围小，都快手动对拍了也不知道

哪儿错了。。。。

我们定义w[i]代表深度<=i的严格n元树的个数

那么最后w[d]-w[d-1]就是答案

那么对于w[i]，我们由w[i-1]递推来，

我们考虑新加一个根节点，然后根节点有n个子节点，每个子节点都可以建一颗深度<=i-1的树，那么每个

子节点都有w[i-1]种选法，那么n个子节点就有w[i-1]^n选法，再加上都不选，就是深度为0的情况

那么w[i]:=（w[i-1]^n）+1；

//By BLADEVIL
var
    w                           :array[-..] of ansistring;
    n, d                        :longint;
    a, b, c                     :array[..] of int64;

function mul(s1,s2:ansistring):ansistring;
var
    i, j                        :longint;
    len1, len2                  :longint;
    s                           :ansistring;
begin
    len1:=length(s1);
    len2:=length(s2);
    fillchar(c,sizeof(c),);
    fillchar(a,sizeof(a),);
    fillchar(b,sizeof(b),);
    for i:= to len1 do a[(len1-i) div +]:=a[(len1-i) div +]*+ord(s1[i])-;
    for i:= to len2 do b[(len2-i) div +]:=b[(len2-i) div +]*+ord(s2[i])-;
    len1:=(len1+) div ;
    len2:=(len2+) div ;
    for i:= to len1 do
        for j:= to len2 do
        begin
            c[i+j-]:=c[i+j-]+a[i]*b[j];
            c[i+j]:=c[i+j]+c[i+j-] div ;
            c[i+j-]:=c[i+j-] mod ;
        end;
    mul:='';
    len1:=len1+len2+;
    for i:=len1 downto  do
    begin
        if c[i]< then mul:=mul+'';
        if c[i]< then mul:=mul+'';
        if c[i]< then mul:=mul+'';
        str(c[i],s);
        mul:=mul+s;
    end;
    while (mul[]='') and (length(mul)>) do delete(mul,,);
end;

function mi(x:ansistring):ansistring;
var
    p                           :longint;
    ans, sum                    :ansistring;
begin
    ans:='';
    sum:=x;
    p:=n;
    while p<> do
    begin
        if p mod = then ans:=mul(ans,sum);
        p:=p div ;
        sum:=mul(sum,sum);
    end;
    mi:=ans;
end;

function inc(x:ansistring):ansistring;
var
    len                         :longint;
    i                           :longint;
    s                           :ansistring;

begin
    len:=length(x);
    for i:= to len do c[i]:=ord(x[i])-;
    c[len]:=c[len]+;
    for i:=len downto  do
    begin
        c[i-]:=c[i-]+c[i] div ;
        c[i]:=c[i] mod ;
    end;
    inc:='';
    len:=len;
    for i:= to len do
    begin
        str(c[i],s);
        inc:=inc+s;
    end;
    while (inc[]='') and (length(inc)>) do delete(inc,,);
end;

function jian(s1,s2:ansistring):ansistring;
var
    i                           :longint;
    len1, len2                  :longint;
    s                           :ansistring;
begin
    len1:=length(s1);
    len2:=length(s2);
    fillchar(c,sizeof(c),);
    for i:= to len1 do a[len1-i+]:=ord(s1[i])-;
    for i:= to len2 do b[len2-i+]:=ord(s2[i])-;
    for i:= to len1 do c[i]:=a[i]-b[i];
    for i:= to len1 do
        if c[i]< then
        begin
            c[i]:=c[i]+;
            c[i+]:=c[i+]-;
        end;
    jian:='';
    for i:=len1 downto  do
    begin
        str(c[i],s);
        jian:=jian+s;
    end;
    while (jian[]='') and (length(jian)>) do delete(jian,,);
end;

procedure main;
var
    i                           :longint;
begin
    readln(n,d);
    if d= then
    begin
        writeln();
        exit;
    end;
    w[]:='';
    for i:= to d do w[i]:=inc(mi(w[i-]));
    writeln(jian(w[d],w[d-]));
end;

begin
    main;
end.