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6:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,A[100],i,dem=0;
cin>>n;
for (int i=1; i<=n; i++) cin>>A[i];
for (int i=1;i<=n; i++)
if (A[i]%2!=0) dem++;
cout<<dem;
return 0;
}
5:
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long n,nn=1e6,A[1000];
cin>>n;
for (int i=1; i<=n; i++) cin>>A[i];
for (int i=1; i<=n; i++)
nn=min(nn,A[i]);
for (int i=1; i<=n; i++)
if (nn==A[i]) cout<<i<<" ";
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int a[100][100],i,j,m,n;
cout<<"Nhap so dong cua mang:"; cin>>n;
cout<<"Nhap so cot cua mang:"; cin>>m;
for (i=1; i<=n; i++)
for (j=1; j<=m; j++)
{
cout<<"A["<<i<<","<<j<<"]="; cin>>a[i][j];
}
for (i=1; i<=n; i++)
{
for (j=1; j<=m; j++)
cout<<a[i][j]<<" ";
cout<<endl;
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
long long a[1000],i,n,k,dem;
int main()
{
cin>>n;
for (i=1; i<=n; i++) cin>>a[i];
cin>>k;
dem=0;
for (i=1; i<=n; i++)
if (a[i]==k) dem++;
cout<<dem;
return 0;
}
#include <bits/stdc++.h>
using namespace std;
long long a[1000],i,n;
int main()
{
cin>>n;
for (i=1; i<=n; i++) cin>>a[i];
for (i=1; i<=n; i++) cout<<a[i]<<" ";
return 0;
}
#include <bits/stdc++.h>
using namespace std;
long long a[1000],i,n,nn;
int main()
{
cin>>n;
for (i=1; i<=n; i++) cin>>a[i];
nn=a[1];
for (i=1; i<=n; i++) nn=min(nn,a[i]);
cout<<nn<<endl;
for (i=1; i<=n; i++) if (nn==a[i]) cout<<i<<" ";
return 0;
}
#include <bits/stdc++.h>
using namespace std;
long long a[1000],i,n,t,dem;
int main()
{
cin>>n;
for (i=1; i<=n; i++) cin>>a[i];
for (i=1; i<=n; i++) cout<<a[i]<<" ";
cout<<endl;
dem=0;
t=0;
for (i=1; i<=n; i++)
if (a[i]>0)
{
dem++;
t=t+sqrt(a[i]);
}
cout<<t;
return 0;
}
Câu 1:
var a:array[1..100]of integer;
Câu 2:
uses crt;
var a:array[1..10]of integer;
i:integer;
begin
clrscr;
for i:=1 to 10 do
begin
write('A[',i,']='); readln(a[i]);
end;
for i:=1 to 10 do
write(a[i]:4);
readln;
end.
Câu 3:
uses crt;
var a:array[1..15]of integer;
i,n,t:integer;
begin
clrscr;
repeat
write('Nhap n='); readln(n);
until (0<n) and (n<=15);
for i:=1 to n do
begin
repeat
write('A[',i,']='); readln(a[i]);
until a[i]<=100;
end;
writeln('Mang da nhap la: ');
for i:=1 to n do
write(a[i]:4);
writeln;
t:=0;
for i:=1 to n do
if a[i] mod 3=0 then t:=t+a[i];
writeln('Tong cac phan tu chia het cho 3 la: ',t);
writeln('Cac so chan o vi tri le trong day la: ');
for i:=1 to n do
if (a[i] mod 2=0) and (i mod 2=1) then write(a[i]:4);
readln;
end.
Câu 1:
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('Nhap n='); readln(n);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
t:=0;
for i:=1 to n do
t:=t+a[i];
writeln('Tong cac so trong mang la: ',t);
readln;
end.
Câu 2:
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('Nhap n='); readln(n);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
t:=0;
for i:=1 to n do
if a[i] mod 2=0 then t:=t+a[i];
writeln('Tong cac so chan la: ',t);
readln;
end.