bzoj 1211 prufer编码+排列组合

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题意：n个点，允许任意两点之间连边，给定每个点的最终度数，求所有满足要求的树的个数

bzoj 1005的简化版，关于prufer序列及答案化简请戳（我才不说我是懒得写了呢...）

因为所有点都有要求，注意特判不合法的情况:

（1）sigma（d[i]-1)<>n-2

（2） (d[i]=0) && (n<>1)

var
     n,x,tot            :longint;
     ans                :int64;
     i                  :longint;
     prime              :array[0..110] of longint;
     vis                :array[0..210] of boolean;
     num                :array[0..110] of longint;

procedure pre_do;
var
        i,j:longint;
begin
   for i:=2 to n do
   begin
      if not vis[i] then
      begin
         inc(prime[0]);
         prime[prime[0]]:=i;
      end;
      for j:=1 to prime[0] do
        if i*prime[j]>n then break else
        begin
           vis[i*prime[j]]:=true;
           if i mod prime[j]=0 then break;
        end;
   end;
end;

procedure calcdown(x:longint);
var
        tt,i:longint;
begin
   for i:=1 to prime[0] do
   begin
      if prime[i]>x then exit;
      tt:=prime[i];
      while tt<=x do
      begin
         dec(num[i],x div tt);
         tt:=tt*prime[i];
      end;
   end;
end;

procedure calcup(x:longint);
var
        tt,i:longint;
begin
   for i:=1 to prime[0] do
   begin
      if prime[i]>x then exit;
      tt:=prime[i];
      while tt<=x do
      begin
         inc(num[i],x div tt);
         tt:=tt*prime[i];
      end;
   end;
end;

function qp(a,b:longint):int64;
var
        ans:int64;
begin
   ans:=1;
   while b>0 do
   begin
      if b and 1=1 then ans:=ans*a;
      a:=a*a;
      b:=b>>1;
   end;
   exit(ans);
end;

begin
   read(n);
   pre_do;
   for i:=1 to n do
   begin
      read(x);
      if (x=0) and (n<>1) then
      begin
         writeln(0); exit;
      end;
      inc(tot,x-1);
      calcdown(x-1);
   end;
   if (tot<>n-2) then
   begin
      writeln(0); exit;
   end;
   calcup(n-2);
   ans:=1;
   for i:=1 to prime[0] do
     if num[i]>0 then ans:=ans*qp(prime[i],num[i]);
   writeln(ans);
end.

——by Eirlys