Use an anti-static bag to hold the RAM while installing.
Answer:see explanation
Explanation:
Hello, from your question I see that you need a program which based on countNum's value prints ready then all the numbers to 1 and finally go!, I'd appreciate if you provided information about the numbers after the example you gave as I do not know if they are to be printed too.
The solution I provide includes the countdown to 1 printing newline after each number and text. Try it out!
#include <stdio.h>
int main(void)
{
int countNum;
int i;
countNum = 22;
for ( i = countNum; i > 0; i--) {
if (i == countNum){
printf("Ready!\n");
}
printf("%d\n",i);
}
printf("Go!\n");
return 0;
}
Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
Answer: Accept only one at a time, all on a slide, or all changes in a presentation.
Explanation:
Took the review
Answer:
const MAXNR=150;
let candidates = {};
for(let i=2; i<=MAXNR; i++) {
candidates[i] = true;
}
for(let p=2; p <= MAXNR; p++) {
if (candidates[p]) {
process.stdout.write(`${p} `);
// Now flag all multiples of p as false
i=2;
while(p*i <= MAXNR) {
candidates[p*i] = false;
i++;
}
}
}
