Answer:
Don’t know lol but look it up on g00gle and she will have the answr
Explanation:
Answer:
- public class Main {
-
- public static void main (String [] args) {
- int[][] myArray = {{1,5,6}, {7, 9, 2}};
- fixArray(myArray, 1, 2, 12);
-
- System.out.println(myArray[1][2]);
- }
-
-
- private static void fixArray(int[][] array, int row, int col, int value){
- array[row][col] = value;
- }
- }
Explanation:
The solution code is written in Java.
Firstly, create the method fixArray with that takes four inputs, array, row, col and value (Line 11). Within the method body, use row and col as index to address a particular element from array and set the input value to it (Line 12).
Next, we test the method in the main program using a sample array (Line 4) and we try to change the row-1 and col-2 element from 2 to 12 (Line 5).
The print statement in Line 7 will display 12 in console.
Hello there,
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Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
Answer:
b
Explanation:
if you look at the end of the code it brings c back but for it to do that it has to make a newline.
