Solution :
int f(int x){
;
}*/
int f(
){
int stack[2] = {
};
;
;
;
R1 = R1 * C; /*R1 = 
*/
;
* R2; /* R0 = bx */
; /*R0 = 
*/
;
; /*R0 = a+bx+cx^2 */
![\text{R1 = stack[0];}](https://tex.z-dn.net/?f=%5Ctext%7BR1%20%3D%20stack%5B0%5D%3B%7D)
![\text{ R2 = stack[1];}](https://tex.z-dn.net/?f=%5Ctext%7B%20R2%20%3D%20stack%5B1%5D%3B%7D)
}
/*
int s=0;
{ 
;)
*/
AREA SumSquares, code, readWrite
ENTRY
;loop 
MOV r1, #0 ; s = 0
Loop 
;calculate i*i
;s = s+ i*i
, #1 ; i = i+1
,#10 ; test for end
Loop ; 
END
B.tablet because it is technically not a computer in traditional sence but a hand held device.
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3. A, (first choice)
4. C (Third choice from top)
Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Answer:
Code is given below and output is attached as an image.
Explanation:
#include <iostream>
#include <fstream>
using namespace std;
bool isPalindrome(int n)
{
// Find reverse of n
int rev = 0;
for (int i = n; i > 0; i /= 10)
rev = rev * 10 + i % 10;
// If n and rev are same,then n is a palindrome
return (n == rev);
}
int main()
{
int min = 1; // Lower Bound
int max = 200; // Upper Bound
ofstream myfile;
myfile.open("palindrome.txt");
for (int i = min + 1; i < max; i++)
if (isPalindrome(i))
myfile << i << endl;
myfile.close();
return 0;
}
