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
[-
1
..
100
]
of
ansistring;
n, d :longint;
a, b, c :
array
[
0
..
100000
]
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),
0
);
fillchar(a,sizeof(a),
0
);
fillchar(b,sizeof(b),
0
);
for
i:=
1
to
len1
do
a[(len1-i)
div
4
+
1
]:=a[(len1-i)
div
4
+
1
]*
10
+ord(s1[i])-
48
;
for
i:=
1
to
len2
do
b[(len2-i)
div
4
+
1
]:=b[(len2-i)
div
4
+
1
]*
10
+ord(s2[i])-
48
;
len1:
=(len1+
3
)
div
4
;
len2:
=(len2+
3
)
div
4
;
for
i:=
1
to
len1
do
for
j:=
1
to
len2
do
begin
c[i
+j-
1
]:=c[i+j-
1
]+a[i]*
b[j];
c[i
+j]:=c[i+j]+c[i+j-
1
]
div
10000
;
c[i
+j-
1
]:=c[i+j-
1
]
mod
10000
;
end
;
mul:
=
''
;
len1:
=len1+len2+
1
;
for
i:=len1
downto
1
do
begin
if
c[i]<
1000
then
mul:=mul+
'
0
'
;
if
c[i]<
100
then
mul:=mul+
'
0
'
;
if
c[i]<
10
then
mul:=mul+
'
0
'
;
str(c[i],s);
mul:
=mul+
s;
end
;
while
(mul[
1
]=
'
0
'
)
and
(length(mul)>
1
)
do
delete(mul,
1
,
1
);
end
;
function
mi(x:ansistring):ansistring;
var
p :longint;
ans, sum :ansistring;
begin
ans:
=
'
1
'
;
sum:
=
x;
p:
=
n;
while
p<>
0
do
begin
if
p
mod
2
=
1
then
ans:=
mul(ans,sum);
p:
=p
div
2
;
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:=
1
to
len
do
c[i]:=ord(x[i])-
48
;
c[len]:
=c[len]+
1
;
for
i:=len
downto
1
do
begin
c[i
-
1
]:=c[i-
1
]+c[i]
div
10
;
c[i]:
=c[i]
mod
10
;
end
;
inc:
=
''
;
len:
=
len;
for
i:=
0
to
len
do
begin
str(c[i],s);
inc:
=inc+
s;
end
;
while
(inc[
1
]=
'
0
'
)
and
(length(inc)>
1
)
do
delete(inc,
1
,
1
);
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),
0
);
for
i:=
1
to
len1
do
a[len1-i+
1
]:=ord(s1[i])-
48
;
for
i:=
1
to
len2
do
b[len2-i+
1
]:=ord(s2[i])-
48
;
for
i:=
1
to
len1
do
c[i]:=a[i]-
b[i];
for
i:=
1
to
len1
do
if
c[i]<
0
then
begin
c[i]:
=c[i]+
10
;
c[i
+
1
]:=c[i+
1
]-
1
;
end
;
jian:
=
''
;
for
i:=len1
downto
1
do
begin
str(c[i],s);
jian:
=jian+
s;
end
;
while
(jian[
1
]=
'
0
'
)
and
(length(jian)>
1
)
do
delete(jian,
1
,
1
);
end
;
procedure
main;
var
i :longint;
begin
readln(n,d);
if
d=
0
then
begin
writeln(
1
);
exit;
end
;
w[
0
]:=
'
1
'
;
for
i:=
1
to
d
do
w[i]:=inc(mi(w[i-
1
]));
writeln(jian(w[d],w[d
-
1
]));
end
;
begin
main;
end
.
