Các pro chỉ giùm mình cách tính n! , n^n với số n rất lớn với @@

bạn có thể vào thư viện xem cách xử lí số lớn và làm thử bài BIGNUM

mới học viết n!,

Code:

Uses Math,windows; Const fi =''; fo =''; MAXN =100000; CS =1000000000; MAXX =15000000000;   Type num = Qword; snum = longint;   Var a : array [0..MAXN] of num; c2,c5 : longint; sz : snum; dau,cuoi : int64;   function gt : int64; begin exit(gettickcount); end;   Procedure Mul(x: num); var i : snum; begin for i:=1 to sz do a[i]:=a[i]*x; for i:=1 to sz-1 do begin inc(a[i+1],a[i] div CS); a[i]:=a[i] mod CS; end; while a[sz]>CS do begin inc(sz); inc(a[sz],a[sz-1] div CS); a[sz-1]:=a[sz-1] mod CS; end; end;   Procedure G2(var y: num); begin while y mod 2 = 0 do begin y:=y div 2; inc(c2); end; end;   Procedure G5(var y: num); begin while y mod 5 = 0 do begin y:= y div 5; inc(c5); end; end;   Procedure nhap; var n,x,y : num;   begin assign(input,fi);reset(input); a[1]:=1; sz:=1; readln(n); while n>1 do begin x:=n*(n-1); dec(n,2); while (x*n<MAXX) and (n>1) do begin y:=x; G2(y);G5(y); x:=y*n; dec(n); end; mul(x); end; close(input); end;   Procedure xuly; var k : longint; x,y : num; begin k:=min(c5,c2); if k = 0 then exit; if c2>c5 then x:=2 else x:=5; while k<max(c5,c2) do begin y:=1; while (y*x<MAXX) and (k<max(c5,c2)) do begin inc(k); y:=y*x; end; Mul(y); end; end;   Procedure ghi(x: num); var t : num = CS div 10; begin while t>x do begin write(0); t:=t div 10; end; if x>0 then write(x); end;   Procedure xuat; var i : snum; begin assign(output,fo);rewrite(output); write(a[sz]); for i:=sz-1 downto 1 do ghi(a[i]); for i:=1 to min(c2,c5) do write(0); close(output); end;   BEGIN dau:=gt;cuoi:=gt;   {--------------------------------------} NHAP; {--------------------------------------}   writeln('NHAP: ',gt-cuoi); cuoi:=gt;   {--------------------------------------} XULY; {--------------------------------------}   writeln('XULY: ',gt-cuoi); cuoi:=gt;   {--------------------------------------} XUAT; {--------------------------------------}   assign(output,'');rewrite(output); writeln('XUAT: ',gt-cuoi); writeln('Tong thoi gian chay chuong trinh : ',gt-dau); close(output); END.

N! thì chắc phải trâu O(n) thôi. n!=(n-1)!*n

Còn n^n thì mình có thể tách thành
n^n = ( ( n^( n div 2 ) ) ^ 2 ) * ( n^( n mod 2 ) )
còn phần xử lý số lớn thì tuỳ mõi người có cách khác nhau.

compoko viết:

QUOTE:
mới học viết n!, :)

Code:

Uses Math,windows; Const fi =''; fo =''; MAXN =100000; CS =1000000000; MAXX =15000000000;   Type num = Qword; snum = longint;   Var a : array [0..MAXN] of num; c2,c5 : longint; sz : snum; dau,cuoi : int64;   function gt : int64; begin exit(gettickcount); end;   Procedure Mul(x: num); var i : snum; begin for i:=1 to sz do a[i]:=a[i]*x; for i:=1 to sz-1 do begin inc(a[i+1],a[i] div CS); a[i]:=a[i] mod CS; end; while a[sz]>CS do begin inc(sz); inc(a[sz],a[sz-1] div CS); a[sz-1]:=a[sz-1] mod CS; end; end;   Procedure G2(var y: num); begin while y mod 2 = 0 do begin y:=y div 2; inc(c2); end; end;   Procedure G5(var y: num); begin while y mod 5 = 0 do begin y:= y div 5; inc(c5); end; end;   Procedure nhap; var n,x,y : num;   begin assign(input,fi);reset(input); a[1]:=1; sz:=1; readln(n); while n>1 do begin x:=n*(n-1); dec(n,2); while (x*n<MAXX) and (n>1) do begin y:=x; G2(y);G5(y); x:=y*n; dec(n); end; mul(x); end; close(input); end;   Procedure xuly; var k : longint; x,y : num; begin k:=min(c5,c2); if k = 0 then exit; if c2>c5 then x:=2 else x:=5; while k<max(c5,c2) do begin y:=1; while (y*x<MAXX) and (k<max(c5,c2)) do begin inc(k); y:=y*x; end; Mul(y); end; end;   Procedure ghi(x: num); var t : num = CS div 10; begin while t>x do begin write(0); t:=t div 10; end; if x>0 then write(x); end;   Procedure xuat; var i : snum; begin assign(output,fo);rewrite(output); write(a[sz]); for i:=sz-1 downto 1 do ghi(a[i]); for i:=1 to min(c2,c5) do write(0); close(output); end;   BEGIN dau:=gt;cuoi:=gt;   {--------------------------------------} NHAP; {--------------------------------------}   writeln('NHAP: ',gt-cuoi); cuoi:=gt;   {--------------------------------------} XULY; {--------------------------------------}   writeln('XULY: ',gt-cuoi); cuoi:=gt;   {--------------------------------------} XUAT; {--------------------------------------}   assign(output,'');rewrite(output); writeln('XUAT: ',gt-cuoi); writeln('Tong thoi gian chay chuong trinh : ',gt-dau); close(output); END.

vậy code này dùng cho n khoảng bao nhiêu vậy ạ?

150000 đổ xuống

ờm! mình chỉ hướng dẫn bạn cách xử lí số nguyên lớn thôi nhé! còn phần còn lại bạn tính n! như bình thường thôi:
{$MODE OBJFPC}
program BIGNUM;
type
sn=record
num:longint;
digit:array[0..500] of longint;
end;

const
tfi='BIGNUM.INP';
tfo='BIGNUM.OUT';
coso=100000000;
cs=8;

var
fi,fo:text;


procedure doc(var x:sn);
var i,j,p:longint;
a:array[1..1001] of char;
ch:char;
begin
repeat
read(fi,ch);
until (ch>='0' and (ch<='9';
p:=1; a[1]:=ch;
while not eof(fi) do
begin
read(fi,ch);
if (ch<'0' or (ch>'9' then break;
inc(p);a[p]:=ch;
end;
with x do
begin
num:=-1;
while p>0 do
begin
i:=p-cs+1;
if i<1 then i:=1;
inc(num);
digit[num]:=0;
for j:=i to p do digit[num]:=digit[num]*10+ord(a[j])-48;
p:=i-1;
end;
while (num>0) and (digit[num]=0) do dec(num)
end;
end;

procedure tong(a,b:sn;var c:sn);
var i,nho,tong:longint;
begin
c.num:=a.num;
if c.num<b.num then c.num:=b.num;
for i:=a.num+1 to c.num do a.digit[i]:=0;
for i:=b.num+1 to c.num do b.digit[i]:=0;
nho:=0;
for i:=0 to c.num do
begin
tong:=a.digit[i]+b.digit[i]+nho;
c.digit[i]:=tong mod coso;
nho:=tong div coso;
end;
if nho>0 then
begin
inc(c.num);
c.digit[c.num]:=1;
end;
end;

procedure viet(var x:sn);
var i:longint;
s:AnsiString;
begin
with x do
begin
write(fo,digit[num]);
for i:=num-1 downto 0 do
begin
str(digit[i],s);
while length(s)<cs do s:='0'+s;
write(fo,s);
end;
end;
writeln(fo);
end;

function lonhon(var a,b:sn):boolean;
var i:longint;
begin
if a.num>b.num then exit(true);
if a.num<b.num then exit(false);
i:=a.num;
while (i>=0) and (a.digit[i]=b.digit[i]) do dec(i);
lonhon:=(i>=0) and (a.digit[i]>b.digit[i]);
end;

procedure tru(a,b:sn; var c:sn);
var nho,hieu,i:longint;
begin
c.num:=a.num;
for i:=b.num+1 to c.num do b.digit[i]:=0; nho:=0;
for i:=0 to c.num do
begin
hieu:=a.digit[i]-b.digit[i]-nho;
if hieu<0 then
begin
hieu:=hieu+coso;
nho:=1;
end
else nho:=0;
c.digit[i]:=hieu;
end;
while (c.num>0) and (c.digit[c.num]=0) do dec(c.num);
end;


procedure nhan(a,b:sn; var c:sn);
var tich,nho,s:int64;
i,j:longint;
begin
c.num:=a.num+b.num;
for i:=a.num+1 to c.num do a.digit[i]:=0;
for i:=b.num+1 to c.num do b.digit[i]:=0;
nho:=0;
for i:=0 to c.num do
begin
tich:=nho;
for j:=0 to i do
begin
s:=a.digit[j];
tich:=tich+s*b.digit[i-j];
end;
c.digit[i]:=tich mod coso; nho:=tich div coso;
end;
while nho>0 do
begin
inc(c.num);
c.digit[c.num]:=nho mod coso;
nho:=nho div coso;
end;
while (c.num>0) and (c.digit[c.num]=0) do dec(c.num);
end;


procedure main;
var a,b,c,d,e:sn;
begin
assign(fi,tfi);reset(fi);
assign(fo,tfo);rewrite(fo);
doc(a);doc(b);
tong(a,b,c); viet(c);
if lonhon(b,a) then
begin
tru(b,a,d);
write(fo,'-';viet(d);
end
else
begin
tru(a,b,d);
viet(d);
end;
nhan(a,b,e);viet(e);
close(fi);close(fo);
end;

BEGIN
main;
END.

đây là file .pas

{$MODE OBJFPC}
program BIGNUM;
type
sn=record
num:longint;
digit:array[0..500] of longint;
end;

const
tfi='BIGNUM.INP';
tfo='BIGNUM.OUT';
coso=100000000;
cs=8;

var
fi,fo:text;


procedure doc(var x:sn);
var i,j,p:longint;
a:array[1..1001] of char;
ch:char;
begin
repeat
read(fi,ch);
until (ch>='0' and (ch<='9';
p:=1; a[1]:=ch;
while not eof(fi) do
begin
read(fi,ch);
if (ch<'0' or (ch>'9' then break;
inc(p);a[p]:=ch;
end;
with x do
begin
num:=-1;
while p>0 do
begin
i:=p-cs+1;
if i<1 then i:=1;
inc(num);
digit[num]:=0;
for j:=i to p do digit[num]:=digit[num]*10+ord(a[j])-48;
p:=i-1;
end;
while (num>0) and (digit[num]=0) do dec(num)
end;
end;

procedure tong(a,b:sn;var c:sn);
var i,nho,tong:longint;
begin
c.num:=a.num;
if c.num<b.num then c.num:=b.num;
for i:=a.num+1 to c.num do a.digit[i]:=0;
for i:=b.num+1 to c.num do b.digit[i]:=0;
nho:=0;
for i:=0 to c.num do
begin
tong:=a.digit[i]+b.digit[i]+nho;
c.digit[i]:=tong mod coso;
nho:=tong div coso;
end;
if nho>0 then
begin
inc(c.num);
c.digit[c.num]:=1;
end;
end;

procedure viet(var x:sn);
var i:longint;
s:AnsiString;
begin
with x do
begin
write(fo,digit[num]);
for i:=num-1 downto 0 do
begin
str(digit[i],s);
while length(s)<cs do s:='0'+s;
write(fo,s);
end;
end;
writeln(fo);
end;

function lonhon(var a,b:sn):boolean;
var i:longint;
begin
if a.num>b.num then exit(true);
if a.num<b.num then exit(false);
i:=a.num;
while (i>=0) and (a.digit[i]=b.digit[i]) do dec(i);
lonhon:=(i>=0) and (a.digit[i]>b.digit[i]);
end;

procedure tru(a,b:sn; var c:sn);
var nho,hieu,i:longint;
begin
c.num:=a.num;
for i:=b.num+1 to c.num do b.digit[i]:=0; nho:=0;
for i:=0 to c.num do
begin
hieu:=a.digit[i]-b.digit[i]-nho;
if hieu<0 then
begin
hieu:=hieu+coso;
nho:=1;
end
else nho:=0;
c.digit[i]:=hieu;
end;
while (c.num>0) and (c.digit[c.num]=0) do dec(c.num);
end;


procedure nhan(a,b:sn; var c:sn);
var tich,nho,s:int64;
i,j:longint;
begin
c.num:=a.num+b.num;
for i:=a.num+1 to c.num do a.digit[i]:=0;
for i:=b.num+1 to c.num do b.digit[i]:=0;
nho:=0;
for i:=0 to c.num do
begin
tich:=nho;
for j:=0 to i do
begin
s:=a.digit[j];
tich:=tich+s*b.digit[i-j];
end;
c.digit[i]:=tich mod coso; nho:=tich div coso;
end;
while nho>0 do
begin
inc(c.num);
c.digit[c.num]:=nho mod coso;
nho:=nho div coso;
end;
while (c.num>0) and (c.digit[c.num]=0) do dec(c.num);
end;


procedure main;
var a,b,c,d,e:sn;
begin
assign(fi,tfi);reset(fi);
assign(fo,tfo);rewrite(fo);
doc(a);doc(b);
tong(a,b,c); viet(c);
if lonhon(b,a) then
begin
tru(b,a,d);
write(fo,'-';viet(d);
end
else
begin
tru(a,b,d);
viet(d);
end;
nhan(a,b,e);viet(e);
close(fi);close(fo);
end;

BEGIN
main;
END.