The pseudocode to find the sum of all integers that are multiples of 9, from 1 to 250.
totalSum = 0
for i from 1 to 250{
if i is divided by 9 and remainder is 0{
totalSum = totalSum + i;
}
}
print(totalSum)
in python language the code will be
totalSum = 0
for i in range(1,250):
if i%9==0:
totalSum += i
If you will run the program , the answer would be 3402.
Answer:
Output:
123456
123456
123456
123456
123456
123456
Explanation:
C Code:
#include <stdio.h>
int main() {
int n,i,j;
printf("Gimme a decimal value to use as n:");
scanf("%d",&n);
for(i=0;i<n;i++){
for(j=1;j<=n;j++){
printf("%d",j);
}
printf("\n");
}
return 0;
}
Equivalent assembly program:
.LC0:
.string "Gimme a decimal value to use as n:"
.LC1:
.string "%d"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
lea rax, [rbp-12]
mov rsi, rax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call scanf
mov DWORD PTR [rbp-4], 0
.L5:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-4], eax
jge .L2
mov DWORD PTR [rbp-8], 1
.L4:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-8], eax
jg .L3
mov eax, DWORD PTR [rbp-8]
mov esi, eax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call printf
add DWORD PTR [rbp-8], 1
jmp .L4
.L3:
mov edi, 10
call putchar
add DWORD PTR [rbp-4], 1
jmp .L5
.L2:
mov eax, 0
leave
ret
Input:
6
The program was first written with a c code, anf and subsequently translated to an assembly language.
The intake ports must be open to allow fuel and oxygen to enter cylinder, then closes during compression. Your answer is A.
Answer:
D
Explanation:
Search engine optimization (SEO)
