Thứ Sáu, 17 tháng 8, 2012

Các bài tập Pascal cơ bản


Sau đây là đề...

Bài 1:
 
Nhp vào 1 s nguyên gm 4 ch s
 
-Kim tra tình chn l
 
-kim tra xem có phi là s nguyên t không
 
-kim tra xem có phi là s hoàn ho không
 
 
 
Bài 2:
 
Nhp vào 1 s n nguyên (2<N<n,m<n7)
 
 
 
Bài 16:
 
-Nhp vào mt danh sách sinh viên (h tên,năm sinh,dan tcim 1 im 2 , đim 3 ,gii tính)
 
-hin danh sách sinh viên va nhp dưới dng ct
 
STT  Hoten  Gioitinh  Dantoc  Diem1  Diem2  Diem3  DTB
 
-cho bit có bao nhiu phn trăm sinh viên nam bao nhiu phn trăm n
 
 
 
Bai 17:
 
-Nhp vào mt danh sách sinh viên (h tên,năm sinh,dan tcim 1 im 2 , đim 3 ,gii tính)
 
-hin danh sách sinh viên va nhp dưới dng ct
 
STT  Hoten  Gioitinh  Dantoc  Diem1  Diem2  Diem3  DTB
 
-sp xếp li danh sách sinh viên tăng dn theo đim trung bình (không làm sai lch thông tin)








Sau đây là lời giải ...
Bài 1:
 
program yen1;
 
uses crt;
 
var i,n,k,d:integer;
 
ok:boolean;
 
function chanle(n:integer):boolean;
 
begin
 
ok:=true;
 
if n mod 2<>0 then ok:=false;
 
chanle:=ok;
 
end;
 
function nguyento(n:integer):boolean;
 
begin
 
ok:=true;
 
d:=0;
 
for i:=1 to n-1 do
 
if n mod i=0 then d:=d+1;
 
if d<>1 then ok:=false;
 
nguyento:=ok;
 
end;
 
function chinhphuong(n:integer):boolean;
 
begin
 
ok:=true;
 
k:=trunc(sqrt(n));
 
if sqr(k)<>n then ok:=false;
 
chinhphuong:=ok;
 
end;
 
begin
 
clrscr;
 
write(' Nhap n=');readln(n);
 
if chanle(n) then writeln(' So vua nhap la chan')
 
else writeln(' So vua nhap la so le');
 
if nguyento(n) then writeln(' So vua nhap la nguyen to')
 
else writeln(' So vua nhap khong phai la so nguyen to');
 
if chinhphuong(n) then writeln(' So vua nhap la so chinh phuong')
 
else writeln(' So vua nhap khong la so chinh phuong');
 
readln;
 
end.


Bài 2:
 
program yen2;
 
uses crt;
 
var a:array [ 1..20] of byte;
 
i,n,d,j,k:integer;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
begin
 
write(' a[',i,']=');readln(a[i]);
 
end;
 
end;
 
procedure nguyento;
 
begin
 
write(' Day so cac so nguyen to:');
 
k:=0;
 
for i:=1 to n do
 
begin
 
d:=0;
 
for j:=1 to a[i]-1 do
 
if a[i] mod j=0 then d:=d+1;
 
if d=1 then begin
 
write(' ',a[i]);
 
k:=k+1;
 
end;
 
end;
 
if k=0 then write(' khong co so nao')
 
else begin
 
writeln;
 
writeln(' Trong day co ',k,' so nguyen to');
 
end;
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n=');readln(n);
 
until (2<N)AND(N< readln; nguyento; nhap;>
 
 
 
 
 
 
Bài 3:
 
program yen3;
 
uses crt;
 
var a:array [1..20] of integer;
 
b:array [1..20] of integer;
 
i,n,max,j,tg:integer;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
begin
 
write(' a[',i,']=');readln(a[i]);
 
end;
 
end;
 
procedure timmax;
 
begin
 
max:=a[1];
 
for i:=2 to n do
 
if a[i]>max then max:=a[i];
 
j:=1;
 
for i:=1 to n do
 
if a[i]=max then begin
 
b[j]:=i;
 
j:=j+1;
 
end;
 
write(' So lon nhat trong day la ',max,' o vi tri thu');
 
for i:=1 to j-1 do
 
write(', ',b[i]);
 
writeln;
 
end;
 
procedure tang;
 
begin
 
for i:=1 to n-1 do
 
for j:=i+1 to n do
 
if a[i]>a[j] then begin
 
tg:=a[i];
 
a[i]:=a[j];
 
a[j]:=tg;
 
end;
 
write(' Day duoc sap xep lai la:');
 
for i:=1 to n do
 
write(' ',a[i]);
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n=');readln(n);
 
until (2<N)AND(N< readln; nhap; tang; timmax;>
 
 
 
 
 
 
Bài 4:
 
program yen4;
 
uses crt;
 
var n,i,d,k,j:integer;
 
procedure hoanhao;
 
begin
 
d:=0;
 
for i:=1 to n-1 do
 
if n mod i=0 then d:=d+i;
 
if d=n then writeln(' So vua nhap la so hoan hao')
 
else writeln(' So vua nhap khong la so hoan hao');
 
end;
 
procedure vietlen;
 
begin
 
write(' Day so hoan hao tu 1-->2000:');
 
k:=0;
 
for i:=1 to 2000 do
 
begin
 
d:=0;
 
for j:=1 to i-1 do
 
if i mod j=0 then d:=d+j;
 
if d=i then begin
 
write(' ',i);
 
k:=k+1;
 
end;
 
end;
 
writeln;
 
writeln(' Trong day 1-->2000 co ',k,' so hoan hao');
 
end;
 
begin
 
clrscr;
 
write(' Nhap n=');readln(n);
 
hoanhao;
 
vietlen;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 5:
 
program yen5;
 
uses crt;
 
var a:array [1..20] of real;
 
b:array [1..20] of integer;
 
i,n,d,j:integer;
 
s,m:real;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
begin
 
write(' a[',i,']=');readln(a[i]);
 
end;
 
end;
 
function tbc:real;
 
begin
 
d:=0;s:=0;
 
for i:=1 to n do
 
if a[i]>0 then begin
 
s:=s+a[i];
 
d:=d+1;
 
end;
 
tbc:=(s/d);
 
end;
 
procedure kiemtra;
 
begin
 
d:=0;j:=1;
 
for i:=1 to n do
 
if m=a[i] then begin
 
b[j]:=i;
 
d:=d+1;
 
j:=j+1;
 
end;
 
if d=0 then writeln(' Trong mang khong co so nao trung trung binh cong')
 
else begin
 
write(' Trong day co ',d,' so trung voi trung binh cong o vi tri');
 
for i:=1 to j-1 do
 
write(', ',b[i]);
 
end;
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n=');readln(n);
 
until (2<N)AND(N< readln; nhap; kiemtra; :',m:2:3); cong bing Trung writeln(' m:="tbc;">
 
 
 
 
 
 
Bài 6:
 
program yen6;
 
uses crt;
 
var st:string;
 
i,d:integer;
 
ch:char;
 
function demkitu:integer;
 
begin
 
d:=0;
 
for i:=1 to length(st) do
 
if (st[i]='c')or(st[i]='C') then d:=d+1;
 
demkitu:=d;
 
end;
 
function demtu:integer;
 
begin
 
d:=1;
 
for i:=1 to length(st) do
 
if (st[i]<>' ')and(st[i+1]=' ') then d:=d+1;
 
demtu:=d;
 
end;
 
procedure chuanhoa;
 
begin
 
while st[1]=' ' do
 
delete(st,1,1);
 
while st[length(st)]=' ' do
 
delete(st,length(st),1);i:=1;
 
repeat
 
if (st[i]=' ')and(st[i+1]=' ') then delete(st,i,1) else i:=i+1;
 
until i>length(st);
 
end;
 
function dem:integer;
 
begin
 
d:=0;
 
for i:=1 to length(st) do
 
if st[i]=ch then d:=d+1;
 
dem:=d;
 
end;
 
begin
 
clrscr;
 
write(' Nhap xau:');readln(st);
 
write(' Nhap ki tu:');readln(ch);
 
chuanhoa;
 
writeln(' Trong xau co ',demkitu,' ki tu C');
 
writeln(' Trong xau co ',demtu,' tu');
 
writeln(' Ki tu ',ch,' vua nhap co mat ',dem,' lan trong xau');
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 7:
 
program yen7;
 
uses crt;
 
var a:array[1..10,1..10] of integer;
 
i,j,n,m:integer;
 
procedure nhap;
 
begin
 
for i:=1 to m do
 
for j:=1 to n do
 
begin
 
write(' a[',i,',',j,']=');readln(a[i,j]);
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln(' Ma tran vua nhap la:');
 
for i:=1 to m do
 
begin
 
for j:=1 to n do
 
write(' ',a[i,j]:4);
 
writeln;
 
end;
 
end;
 
begin
 
clrscr;
 
write(' Nhap m,n:');readln(m,n);
 
nhap;
 
hienthi;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 8:
 
program yen8;
 
uses crt;
 
var a:array [1..100] of integer;
 
n,i,m,d,s:integer;
 
procedure fibonaci;
 
begin
 
if n=1 then a[1]:=1
 
else begin
 
i:=2;s:=1;a[1]:=1;
 
repeat
 
a[i]:=s;
 
i:=i+1;
 
s:=a[i-1]+a[i-2];
 
 
 
until i>n;
 
end;
 
write(' Day so fibonaci la:');
 
for i:=1 to n do
 
write(' ',a[i]);
 
writeln;
 
end;
 
begin
 
clrscr;
 
write(' Nhap so de tinh day fibonaci:');readln(n);
 
write(' Nhap so de kiem tra xem co trong day fibonaci khong:');readln(m);
 
fibonaci;
 
d:=0;
 
for i:=1 to n do
 
if a[i]=m then d:=d+1;
 
if d=1 then writeln(' So de kiem tra co mat trong day fibonaci')
 
else writeln(' So kiem tra khong co mat trong day fibonaci');
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 9:
 
program yen9;
 
uses crt;
 
var tu:string;
 
i:integer;
 
function doixung(tu1:string):boolean;
 
var ok:boolean;
 
begin
 
ok:=true;
 
for i:=1 to (length(tu1) div 2) do
 
if tu1[i]<>tu1[length(tu1)-i+1] then begin
 
ok:=false;
 
break;
 
end;
 
doixung:=ok;
 
end;
 
begin
 
clrscr;
 
write(' Nhap tu:');readln(tu);
 
if doixung(tu) then writeln(' Tu vua nhap doi xung')
 
else writeln(' Tu vua nhap khong doi xung');
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 10:
 
program yen10;
 
uses crt;
 
var a:array [1..10,1..10] of integer;
 
b:array [1..10] of integer;
 
i,j,n,m,max:integer;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
write(' a[',i,',',j,']=');readln(a[i,j]);
 
end;
 
end;
 
procedure xaydung;
 
begin
 
for j:=1 to m do
 
begin
 
max:=a[1,j];
 
for i:=2 to n do
 
if a[j,i]>max then max:=a[j,i];
 
b[j]:=max;
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln(' Ma tran vua nhap la:');
 
for i:=1 to n do
 
begin
 
for j:=1 to m do
 
write(' ',a[i,j]:4);
 
writeln;
 
end;
 
writeln;
 
write(' Day gom cac so max la:');
 
for i:=1 to m do
 
write(' ',b[i]);
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n,m:');readln(n,m);
 
until (2<N)AND(M< readln; nhap; hienthi; xaydung;>
 
 
 
 
 
 
Bài 11:
 
program yen11;
 
uses crt;
 
var  a:array [1..10,1..10] of integer;
 
b:array [1..100] of integer;
 
i,j,n,m,d,k,t:integer;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
write(' a[',i,',',j,']=');readln(a[i,j]);
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln(' Ma tran vua nhap la:');
 
for i:=1 to n do
 
begin
 
for j:=1 to m do
 
write(' ',a[i,j]:4);
 
writeln;
 
end;
 
writeln;
 
end;
 
procedure nguyento;
 
begin
 
k:=1;
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
d:=0;
 
for t:=1 to a[i,j]-1 do
 
if a[i,j] mod t=0 then d:=d+1;
 
if d=1 then begin
 
b[k]:=a[i,j];
 
k:=k+1;
 
end;
 
end;
 
writeln;writeln;
 
writeln(' Day so nguyen to la:');writeln;
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
for t:=1 to k-1 do
 
if b[t]=a[i,j] then begin
 
write(' ',b[t]:8);
 
writeln(' :hang ',i,' cot ',j,' ');
 
end;
 
end;
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n,m:');readln(n,m);
 
until (2<N)AND(M< readln; nguyento; nhap; hienthi;>
 
 
 
 
 
 
Bài 12:
 
program yen12;
 
uses crt;
 
var a:array [1..10,1..10] of integer;
 
b:array [1..100] of integer;
 
i,j,n,m,k,tg:integer;
 
procedure nhap;
 
begin
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
write(' a[',i,',',j,']=');readln(a[i,j]);
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln(' Ma tran vua nhap la:');
 
for i:=1 to n do
 
begin
 
for j:=1 to m do
 
write(' ',a[i,j]:4);
 
writeln;
 
end;
 
writeln;
 
end;
 
procedure sapxep;
 
begin
 
k:=1;
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
b[k]:=a[i,j];
 
k:=k+1;
 
end;
 
for i:=1 to k-2 do
 
for j:=i+1 to k-1 do
 
if b[i]>b[j] then begin
 
tg:=b[i];
 
b[i]:=b[j];
 
b[j]:=tg;
 
end;
 
k:=1;
 
for i:=1 to n do
 
for j:=1 to m do
 
begin
 
a[i,j]:=b[k];
 
k:=k+1;
 
end;
 
writeln(' Ma tran duoc sap xep lai la:');
 
for i:=1 to n do
 
begin
 
for j:=1 to m do
 
write(' ',a[i,j]:4);
 
writeln;
 
end;
 
end;
 
begin
 
clrscr;
 
repeat
 
write(' Nhap n,m:');readln(n,m);
 
until (2<N)AND(M< readln; nhap; hienthi; sapxep;>
 
 
 
 
 
 
Bài 13:
 
program yen13;
 
uses crt;
 
var s1,s2:string;
 
i,j,d,k,t:integer;
 
procedure kiemtra;
 
begin
 
d:=0;
 
for i:=1 to length(s2) do
 
if s2[i]=s1[1] then begin
 
k:=2;t:=1;
 
for j:=i+1 to (i+length(s1)-1) do
 
if s2[j]=s1[k] then begin
 
k:=k+1;
 
t:=t+1;
 
end;
 
if t=length(s1) then d:=d+1;
 
end;
 
if d>0 then begin
 
writeln(' Xau s1 co xuat hien trogn xau s2');
 
writeln(' Xau s1 xuat hien ',d,' lan trong xau s2');
 
end
 
else writeln(' Xau s1 khong xuat hien trong xau s2');
 
end;
 
begin
 
clrscr;
 
write(' Nhap xau s1:');readln(s1);
 
write(' Nhap xau s2:');readln(s2);
 
kiemtra;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 14:
 
program yen14;
 
uses crt;
 
type tap=set of char;
 
var ch:char;
 
st:string;
 
chucai:tap;
 
i,d:integer;
 
procedure kiemtra;
 
begin
 
if ch in chucai then writeln(' Ki tu ',ch,' co trong bang chu cai')
 
else writeln(' Ki tu ',ch,' khong co trong bang chu cai');
 
d:=0;
 
for i:=1 to length(st) do
 
if st[i] in chucai then d:=d+1;
 
writeln(' Trong xau co ',length(st)-d,' ki tu khong la chu cai');
 
end;
 
begin
 
clrscr;
 
write(' Nhap ki tu:');readln(ch);
 
write(' Nhap xau:');readln(st);
 
chucai:=['a'..'z','A'..'Z'];
 
kiemtra;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 15:
 
program yen15;
 
uses crt;
 
type sinhvien=record
 
hoten,dantoc,gioitinh:string;
 
namsinh,diem1,diem2,diem3:integer;
 
dtb:real;
 
end;
 
mang=array [1..10] of sinhvien;
 
var sv:mang;
 
i,n,d:integer;
 
procedure nhap;
 
begin
 
clrscr;
 
writeln('                                                NHAP THONG TIN CHO SINH VIEN');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
write(' Ho ten:');readln(hoten);
 
write(' Gioi tinh:');readln(gioitinh);
 
write(' Dan toc:');readln(dantoc);
 
write(' Diem1,Diem2,Diem3:');readln(diem1,diem2,diem3);
 
dtb:=(diem1+diem2+diem3)/3;
 
writeln;
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln;writeln;
 
writeln('                                               DANH SACH SINH VIEN VUA NHAP');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
writeln(' Ho ten:',hoten);
 
writeln(' Nam sinh:',namsinh);
 
writeln(' Diem TB:',dtb:2:2);
 
writeln;writeln;
 
end;
 
end;
 
procedure sapxep;
 
begin
 
writeln('                                               DANH SACH SINH VIEN DAT LOAI KHA');
 
writeln;
 
d:=0;
 
for i:=1 to n do
 
if sv[i].dtb>7 then begin
 
writeln(' Ho ten:',sv[i].hoten);
 
writeln(' Diem TB:',sv[i].dtb:2:3);
 
writeln;writeln;
 
d:=d+1;
 
end;
 
if d=0 then writeln('                                                   ( khong co sinh vien nao )');
 
end;
 
begin
 
clrscr;
 
write(' Nhap n=');readln(n);
 
nhap;
 
hienthi;
 
sapxep;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 16:
 
program yen16;
 
uses crt;
 
type sinhvien=record
 
hoten,dantoc,gioitinh:string;
 
namsinh,diem1,diem2,diem3:integer;
 
dtb:real;
 
end;
 
mang=array [1..10] of sinhvien;
 
var sv:mang;
 
i,n,d:integer;
 
p:real;
 
procedure nhap;
 
begin
 
clrscr;
 
writeln('                                                NHAP THONG TIN CHO SINH VIEN');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
write(' Ho ten:');readln(hoten);
 
write(' Gioi tinh:');readln(gioitinh);
 
write(' Dan toc:');readln(dantoc);
 
write(' Diem1,Diem2,Diem3:');readln(diem1,diem2,diem3);
 
dtb:=(diem1+diem2+diem3)/3;
 
writeln;
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln;writeln;
 
writeln('                                               DANH SACH SINH VIEN VUA NHAP');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
writeln(' Ho ten:',hoten);
 
writeln(' Nam sinh:',namsinh);
 
writeln(' Diem TB:',dtb:2:2);
 
writeln;writeln;
 
end;
 
end;
 
procedure phantram;
 
begin
 
writeln;writeln;
 
writeln('                                                              TY LE NAM NU');
 
d:=0;
 
for i:=1 to n do
 
if sv[i].gioitinh='nam' then d:=d+1;
 
p:=(d/n)*100;
 
writeln('                               Trong danh sach co ',p:2:2,' % la sinh vien nam');
 
writeln('                               Trong danh sach co ',(100-p):2:2,' % la sinh vien nu');
 
end;
 
begin
 
clrscr;
 
write(' Nhap n=');readln(n);
 
nhap;
 
hienthi;
 
phantram;
 
readln;
 
end.
 
 
 
 
 
 
 
Bài 17:
 
program yen17;
 
uses crt;
 
type sinhvien=record
 
hoten,dantoc,gioitinh:string;
 
namsinh,diem1,diem2,diem3:integer;
 
dtb:real;
 
end;
 
mang=array [1..10] of sinhvien;
 
var sv:mang;
 
i,n,j:integer;
 
procedure nhap;
 
begin
 
clrscr;
 
writeln('                                                NHAP THONG TIN CHO SINH VIEN');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
write(' Ho ten:');readln(hoten);
 
write(' Gioi tinh:');readln(gioitinh);
 
write(' Dan toc:');readln(dantoc);
 
write(' Diem1,Diem2,Diem3:');readln(diem1,diem2,diem3);
 
dtb:=(diem1+diem2+diem3)/3;
 
writeln;
 
end;
 
end;
 
procedure hienthi;
 
begin
 
writeln;writeln;
SACH 
writeln('                                                DANH SINH VIEN VUA NHAP');
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
writeln(' Ho ten:',hoten);
 
writeln(' Nam sinh:',namsinh);
 
writeln(' Diem TB:',dtb:2:2);
 
writeln;writeln;
 
end;
 
end;
 
procedure sapxep;
 
var tg:sinhvien;
 
begin
 
writeln;writeln;
 
writeln('                                        DANH SACH SINH VIEN DUOC SAP XEP LAI LA');
 
for i:=1 to n-1 do
 
for j:=i+1 to n do
 
if sv[i].dtb>sv[j].dtb then begin
 
tg:=sv[i];
 
sv[i]:=sv[j];
 
sv[j]:=tg;
 
end;
 
for i:=1 to n do
 
with sv[i] do
 
begin
 
writeln(' Ho ten:',hoten);
 
writeln(' Diem TB:',dtb:2:2);
 
writeln;
 
end;
 
end;
 
begin
 
clrscr;
 
write(' Nhap n=');readln(n);
 
nhap;
 
hienthi;
 
sapxep;
 
readln;
 
end.
 
 
 
 

program baitap;
 
uses crt;
 
var x:array[1..8]of integer;
 
a:array [1..8]of boolean;
 
b:array[-7..7]of boolean;
 
c:array[2..16]of boolean;
 
k,i:integer;
 
procedure khoitao;
 
begin
 
for i:=1 to 8 do
 
a[i]:=true;
 
for i:=-7 to 7 do
 
b[i]:=true;
 
for i:=2 to 16 do
 
c[i]:=true;
 
end;
 
procedure ghinhan;
 
begin
 
for i:=1 to 8 do
 
write(x[i]);
 
writeln;
 
k:=k+1;
 
end;
 
procedure try(i:integer );
 
var j:integer;
 
begin
 
for j:=1 to 8 do
 
if a[j] and b[i-j] and c[i+j] then
 
begin
 
x[i]:=j;
 
a[j]:=false;
 
b[i-j]:=false;
 
c[i+j]:=false;
 
if i=8 then begin ghinhan;
 
a[j]:=true;
 
b[i-j]:=true;
 
c[i+j]:=true;
 
end
 
else
 
begin
 
try(i+1);
 
a[j]:=true;
 
b[i-j]:=true;
 
c[i+j]:=true;
 
end;
 
end;
 
end;
 
BEGIN
 
clrscr;
 
khoitao;
 
writeln;
 
writeln;
 
try(1);
 
writeln('co the sap xep duoc ',k,' cach');
 
readln;
 
END.
  • Đề:

    Tính giá trị biểu thức bằng pascal :(

    1-x2/2!+x4/4!-x6/6!...



    Quote
1-x2/2!+x4/4!-x6/6!...

Đây là x^2 hay là 2*x .





Đây là x^2 hay là 2*x .


Cái này Em cứ copy vào notepad rùi đổi đuôi thành .Pas là được. Sau đó lưu thành file.


Không có nhận xét nào:

Đăng nhận xét