I programmed in batch before so I know a lot of things!
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Answer:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Explanation:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Answer:
<em>Given Data:</em>
<em>myBytes BYTE 10h, 20h, 30h, 40h </em>
<em>myWords WORD 3 DUP(?), 2000h </em>
<em>myString BYTE "ABCDE"</em>
<em />
Based on the data that we are given we can conclude that:
(a). a. EAX = 1
b. EAX = 4
c. EAX = 4
d. EAX = 2
e. EAX = 4
f. EAX = 8
g. EAX = 5
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:
Interface
Explanation:
Java does not supports multiple inheritance by class, but we can implement multiple inharitacnce in java using interface, with interface we uses implements keyword.
Example-
Declare interface like below code -
interface parent{
}
and use in child class like below code -
class Child1 implements interface{
}
class Child2 implements interface{
}
