trinh viet chuong nhap vao 1 mang 1 chieu gom n phan tu ( voi n duoc nhap vao tu ban phim ). Hay thuc hien cac cong viec sau:
-Nhap/xuar mang
- Dem co bao nhieu phan tu la so chan trong mang
-Kiem tra xem mang co phai la mang le hay khong ? ( tat ca cac phan tu mang la so le )
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viet chuong trinh con thuc hien nhap vao mang 1 chieu A gom N so nguyen . sau do thuc hienn ct con bo tham so tuong ung la (N=6 , cac phan tu mang lan luot la 2,4,6,3,5,7)
viet chuong trinh nhap vao mang so nguyen a(toi da 100 phan tu) tu ban phim in len man hinh cac phan tu chia het cho K (khong duoc nhap vao tu ban phim khi chay chuong trinh) trong mang
Cho mang 1 chieu gom cac phan tu nguyen. Viet chuong trinh nhap gtri cho tung phan tu. Ddeems va hien ra nhung phan tu co gtri \(⋮\) 3.
uses crt;
var a:array[1..100]of integer;
i,n,dem:integer;
begin
clrscr;
write('Nhap n='); readln(n);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
dem:=0;
for i:=1 to n do
if a[i] mod 3=0 then
begin
inc(dem);
write(a[i]:4);
end;
writeln;
writeln('So phan tu chia het cho 3 la: ',dem);
readln;
end.
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viet chuong trinh nhap vao mang 1 chieu ( n nho hon hoac bang 250), in ra tat ca cac phan tu va tinh tong cac phan tu cua mang
uses crt;
var a:array[1..250]of integer;
i,n,t:integer;
begin
clrscr;
readln(n);
t:=0;
for i:=1 to n do readln(a[i]);
for i:=1 to n do write(a[i]:4);
writeln;
for i:=1 to n do t:=t+a[i];
writeln(t);
readln;
end.
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Viet chuong trinh nhap vao 1 mang be hon 20 phan tu
a/ sap xep mang theo thu tu lon dAN
b/ gia tri trung binh cua cac phan tu
c/ in ra phan tu lon nhat va phan tu nho nhat
D/ IN Ra cac phan tu la so chan
uses crt;
var j,tg,s,max,min,i,n:longint;
a:array[1..19] of longint;
s1:real;
begin
clrscr;
write('Nhap n: '); readln(n);
for i:=1 to n do
begin
write('a[',i,']=');readln(a[i]);
end;
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;
for i:=1 to n do
write(a[i],' ');writeln;
for i:=1 to n do
s:=s+a[i];
s1:=s/n;
writeln(‘Gia tri trung binh cua cac phan tu la: ‘,s1:0:0);
min:=a[1];max:=a[1];
for i:=2 to n do begin
if a[i]>max then max:=a[i];
if a[i]<min then min:=a[i];
write('Phan tu lon nhat la: ',max,' ','Phan tu nho nhat la: ',min); writeln;
for i:=1 to n do
if (a[i]<>0) and (a[i] mod 2=0) then write(a[i],' ');
readln
end.
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viet chuong trinh nhap vao 1 xau Sva 1 ki tu bat ki tu ban phim . kiem tra trong xau vua nhap co bao nhiweu ki tu vua nhap , xoa tat ca ki tu do trong xau S.
MONG MOI NGUOI GIUP DO ;
uses crt;
var s:string;
ktu:char;
i,d,dem:integer;
begin
clrscr;
write('Nhap xau:'); readln(s);
write('Nhap ki tu:'); readln(ktu);
d:=length(s);
dem:=0;
for i:=1 to d do
if s[i]=ktu then inc(dem);
writeln('So ki tu ',ktu,' trong xau ',s,' la: ',dem);
for i:=1 to d do
if s[i]=ktu then delete(s,i,1);
writeln('Xau sau khi xoa cac ki tu ',ktu,' la: ',s);
readln;
end.
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Giúp mình với ạ !
viet chuong trinh nhap mang a bat ky tu ban phim( n nho hon hoac bang 100) tinh va xuat ra mang hinh cac phan tu le hoac chan cua may vua nhap
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('n='); readln(n);
if (n>0) and (n<=100) then
begin
t:=0;
for i:=1 to n do
begin
write('a[',i,']='); readln(a[i]);
if (a[i] mod 2=0) or (a[i] mod 2=1) then t:=t+a[i];
end;
writeln('tong cac phan tu chan hoac le trong day la: ',t);
end
else writeln('vui long nhap lai');
readln;
end.
CAU 1:
A, Nhap vao so nguyen n ; Tinh tong S= 1+2+3+...+n
B, Nhap vao mot mang gom m so nguyen . Tinh tong cac phan tu cua mang do.
GIUP MINH VOI NHA MOI NGUOI ...
A:
uses crt;
var t,i,n:integer;
begin
clrscr;
write('Nhap n='); readln(n);
t:=0;
for i:=1 to n do
t:=t+i;
writeln(t);
readln;
end.
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B:
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('Nhap m='); readln(m);
for i:=1 to m do
begin
write('A[',i,']='); readln(a[i]);
end;
t:=0;
for i:=1 to m do
t:=t+a[i];
writeln(t);
readln;
end.
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viet phuong trinh nhap vao mang a gom N(N<=100) cac so nguyen roi thuc hien cac yeu cau
1. in ra danh sach cac so chia het cho 4
2. in ra mang hinh gia tri lon nhat chia het cho 7 va chi so do
3. xuat ra mang hinh cac phan tu cua mang do theo thu tu giam dan
uses crt;
var a,b:array[1..100]of integer;
n,i,dem,max,tam,j:integer;
begin
clrscr;
repeat
write('Nhap n='); readln(n);
until (0<n) and (n<=100);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
writeln('Cac so chia het cho 4 la: ');
for i:=1 to n do
if a[i] mod 4=0 then write(a[i]:4);
writeln;
dem:=0;
for i:=1 to n do
if a[i] mod 7=0 then
begin
inc(dem);
b[dem]:=a[i];
end;
max:=b[1];
for i:=1 to dem do
if max<b[i] then max:=b[i];
writeln('So lon nhat chia het cho 7 la: ',max);
writeln('Cac chi so cua no trong day A la: ');
for i:=1 to n do
if max=a[i] then write(i:4);
writeln;
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i]<a[j] then
begin
tam:=a[i];
a[i]:=a[j];
a[j]:=tam;
end;
writeln('Day so sau khi sap xep giam dan la: ');
for i:=1 to n do
write(a[i]:4);
readln;
end.
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