Answer:
class Main {
public static void printPattern( int count, int... arr) {
for (int i : arr) {
for(int j=0; j<count; j++)
System.out.printf("%d ", i);
System.out.println();
}
System.out.println("------------------");
}
public static void main(String args[]) {
printPattern(4, 1,2,4);
printPattern(4, 2,3,4);
printPattern(5, 5,4,3);
}
}
Explanation:
Above is a compact implementation.
10008 is the answer to your question
Answer:
Following is the program in C++ Language
#include <iostream> // header file
using namespace std; // namespace std
int main() // main method
{
int n; // variable declaration
cout<<" Please enter the number :";
cin>>n; // Read the number
if(n>0) // check the condition when number is positive
{
cout<<n<<endl<<"The number is Positive"; // Display number
}
else if(n<0) // check the condition when number is Negative
{
cout<<n<<endl<<"The number is Negative";// Display number
}
else // check the condition when number is Zero
{
cout<<n<<endl<<"The number is Zero";// Display number
}
return 0;
}
Output:
Please enter the number:
64
The number is Positive
Explanation:
Following are the description of the program
- Declared a variable "n" of int type.
- Read the value of "n" by user.
- Check the condition of positive number by using if block statement .If n>0 it print the number is positive.
- Check the condition of negative number by using else if block statement If n<0 it print the number is negative.
- Finally if both the above condition is fail it print the message " The number is Zero"
Answer:
d. The trigger is fired more than once.
Explanation:
What would happen in this situation is that the trigger would be fired more than once. This is because the trigger will be fired when the user updates the record. It will also be fired when the process builder is run.
If the trigger fires more than once, this can be problematic for the developer. Therefore, it is better if the trigger fires just once, as this is the time when the present changes can be placed.
