viet phuong trinh tao tep so.txt gom n so nguyen duong chan
Những câu hỏi liên quan
viet phuong trinh su dung bien mang phap 10 phan tu kieu so nguyen in ra mang hinh so chan
viet chuong trinh nhap vao so mguyen n va day so gom n la so nguyen tu ban phim in ra man hinh so lon nhat trong day so nguyen vua nhap
Lời giải :
program hotrotinhoc ;
var a: array[1..32000] of integer ;
i,n,max : integer ;
begin
write('Nhap n='); readln(n);
writeln('Nhap gia tri cua cac phan tu');
for i:= 1 to n do
begin
write('a[',i,']='); readln(a[i]);
end;
max:=a[1];
for i:= 1 to n do if max<a[i] then max:=a[i];
write('So lon nhat la:',max');
readln
end.
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Tim mot so chan lon nhat co 5 chu so ma 3 chu so dau giu nguyen gia tri tu trai sang phai tao thanh mot so chinh phuong va 3 chu so cuoi giu nguyen thu tu tao thanh 1 lap phuong dung
viet chuong trinh nhap so n nguyen duong (n<=100) sau do tinh tong cai so tu 1 den N viet ket qua ra man hinh
var s,i,n:integer;
begin
Write('n = ');
While (n > 100) and (n < 1) then
Begin
write('Nhap sai, nhap lai ');readln(n);
End;
for i:=1 to n do
s:=s+i;
write('tong la: ',s);
readln;
end.
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1) Viet chuong trinh nhap xau S1, tao xau S2 gom tat ca cac chu so cua xau S1
#include <bits/stdc++.h>
using namespace std;
string s1;
int d,i;
int main()
{
getline(cin,s1);
d=s1.length();
for (i=0; i<=d-1; i++)
if ((s1[i]>=48) and (s1[i]<=57)) cout<<s1[i];
return 0;
}
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1)Viet tap hop B gom can so la tich cua dung 3 so Nguyen to co 1 chu so
2) Viet tap hop A gom cac so chan co 4 chu so va co tich cac chu so bang 10
viet chuong trinh nhap 1 so nguyen khac khong tu ban phim,viet ra man hinh cac so tu nhien chan
uses crt;
var n,i:integer;
begin
clrscr;
repeat
write('Nhap n='); readln(n);
until n>0;
write('Cac so chan trong khoang tu 1 toi ',n,' la: ');
for i:=1 to n do
if i mod 2=0 then write(i:4);
readln;
end.
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program so_tu_nhien_chan;
uses crt;
var n,i:integer;
begin
clrscr;
write('Nhap n='); readln(n);
write('Cac so chan nho hon hoac bang ',n,' la: ');
for i:=2 to n do
if i mod 2=0 then write(i:4,' ;');
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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viet 5 so thap phan gom toan cac chu so chan khac nhau va phan nguyen co mot chu so :
1,2345678
2,4357656
3,3456543
4,4963476
5,8632549
đúng thì k nha
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