Answer:
#include <iostream>
using namespace std;
int main() {
int a[4][5];//declaring a matrix of 4 rows and 5 columns.
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
{
if(i==3)//initializing last row as 0.
{
a[i][j]=0;
}
else//initializing last row as 1.
{
a[i][j]=1;
}
}
}
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
cout<<a[i][j]<<" ";//printing the matrix.
cout<<endl;
}
return 0;
}
Output:-
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 0 0 0 0
Explanation:
I have created a matrix of size 4 rows and 5 columns.I have used for loops to fill the array.To fill the last row with 0 i have used if statement.else we are filling it with 1.
Answer:
Device management controls peripheral devices by sending them commands in their proprietary machine language. The software routine that deals with each device is called a "driver," and the OS requires drivers for each of the peripherals attached to the computer.
Explanation:
Answer:
That was a very great story that I totally did NOT read cause it too long 0-0.
Explanation:
To my ferns.....GET ON RN CAUSE I WONLEY T^T
Anyways wuv c'alls and have a good day :3
Answer:
See the code below and the algorithm explanation on the figure.
Explanation:
The explanation in order to get the answer is given on the figure below.
Solving this problem with C. The program is given below:
#include <stdio.h>
int main(void) {
int n, Even=0, Odd=0, Zeros=0;
for (;;) {
printf("\nEnter the value the value that you want to check(remember just integers): ");
//IF we input a non-numeric character the code end;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
Zeros++;
}
else {
if (n % 2) {
Even++;
}
else {
Odd++;
}
}
}
printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);
return 0;
}
