因为读入的数有10^9大，而数列长度只有50000，想到离散化。可离散后，位置发生改变，排个序，在二分来查找。

要找它前方的所有比它小的树的t-1之和，想到了单点更新区间查询，线段树和数状数组都可以维护。由于长度为5，用5个树。完成这些，会发现结果会爆int64，于是加个高精度。看完后，你肯定。

typearr=recordx,y:longint;end;vardp:array [0..50001,1..5] of int64;m,len,n,k:longint;f:array [0..50001] of longint;tree:array [0..50001] of arr;sum:array [0..101] of longint;procedure qsort(l,r:longint);
vari,j,key,key1:longint;temp:arr;
beginif l>=r then exit;i:=l; j:=r;key:=tree[(l+r) shr 1].x;key1:=tree[(l+r) shr 1].y;repeatwhile (tree[i].x<key) or (tree[i].x=key) and (tree[i].y<key1) do inc(i);while (tree[j].x>key) or (tree[j].x=key) and (tree[j].y>key1) do dec(j);if i<=j thenbegintemp:=tree[i]; tree[i]:=tree[j]; tree[j]:=temp;inc(i);dec(j);end;until i>j;qsort(l,j);qsort(i,r);
end;function bit(n:longint):longint;
beginexit(n and -n);
end;procedure jf(n:qword);
vari:longint;a,b:array [0..100] of longint;
beginfillchar(a,sizeof(a),0);fillchar(b,sizeof(b),0);i:=-1;while n>0 dobegininc(i);a[i]:=n mod 10;n:=n div 10;end;i:=-1;while i<100 dobegininc(i);b[i]:=a[i]+sum[i]+b[i];if b[i]>=10 thenbegininc(b[i+1]);b[i]:=b[i] mod 10;end;end;for i:=0 to 99 dosum[i]:=b[i];
end;function count(n,j:longint):int64;
varans:int64;
beginans:=0;while n>0 dobeginans:=ans+dp[n,j];n:=n-bit(n);end;exit(ans);
end;procedure update(n,j:longint;k:int64);
beginwhile n<=m dobegindp[n,j]:=dp[n,j]+k;n:=n+bit(n);end;
end;procedure dpp(n:longint);
vartem:int64;i,j:longint;
beginfillchar(dp,sizeof(dp),0);fillchar(sum,sizeof(sum),0);len:=1;for i:=1 to n dobegintem:=count(f[i]-1,4);jf(tem);for j:=5 downto 2 dobegintem:=count(f[i]-1,j-1);update(f[i],j,tem);end;update(f[i],1,1);end;len:=100;while(sum[len]=0) and (len>0) do dec(len);for i:=len downto 0 dowrite(sum[i]);writeln;
end;procedure main;
vari:longint;
beginwhile not eof dobeginreadln(n);m:=n;for i:=1 to n-1 dobeginread(tree[i].x);tree[i].y:=i;end;readln(tree[n].x);tree[n].y:=n;qsort(1,n);f[tree[1].y]:=1; k:=0;for i:=1 to n dobeginif tree[i].x=tree[i-1].x then f[tree[i].y]:=f[tree[i-1].y] elsebeginf[tree[i].y]:=k+1;inc(k);end;end;dpp(n);end;
end;beginmain;
end.