The sum of factorial

The sum of factorial

"Title description" calculates the s=1! with high precision +2! +3! +...+n! (N≤50)
Which "! "denotes factorial, for example: 5! =5*4*3*2*1.

"Input format" a positive integer n.
"Output format" a positive integer s that represents the result of the calculation.

Train of thought: stripping by heightening the essence Plus

 ProgramAA;varN,l,ls,i:longint; A:Array[1.. +] ofLongint; S:Array[1.. +] ofThe longint;//s array is the next number to be added, which is the change. procedureJC (I:longint);//This process is used to find the S arrayvarJ:longint;begin    ifI=1  Then    beginL:=1; LS:=1; S[LS]:=1;    Exit End; ifI>1  Then    begin         forj:=1  toLs Dos[j]:=s[j]*i;//think about doing this for Mao. forj:=1  toLs Do            ifs[j]>=Ten  Then            beginInc (S[j+1],S[J]Div Ten); S[J]:=S[J]MoD Ten; End; ifs[ls+1]<>0  ThenInc (LS);  while s[ls]>=10  Do        beginInc (LS); Inc (S[ls],s[ls-1]Div Ten); S[ls-1]:=s[ls-1]MoD Ten; End; ifLs>=l Thenl:=ls; End;End;procedureWork ;varI,j,k:longint;begin     fori:=1  toNThe number of do//currently to be added is the factorial of I.     beginJC (i); forj:=1  toL DoInc (A[J],S[J]);//A few steps to add factorial to a forj:=1  toL Do            ifa[j]>=Ten  Then            beginInc (A[j+1],A[J]Div Ten); A[J]:=A[J]MoD Ten; End; ifa[l+1]<>0  ThenInc (L);  whilea[l]>=Ten  Do        beginInc (L); Inc (A[l],a[l-1]Div Ten); A[l-1]:=a[l-1]MoD Ten; End; End;End;beginREADLN (n);    Work  forI:=lDownto 1  Dowrite (a[i]);End.

A little reflection: At the beginning of the 215, because the red Word part of the error, written a array of ...

The sum of factorial