Computers and Technology, 11.06.2021 03:20 karlyisaunicorn

Countdown until matching digits Write a program that takes in an integer in the range 11-100 as input. The output is a countdown starting from the integer, and stopping when both output digits are identical. Ex: If the input is the output is 93 92 91 90 89 88 Ex: If the input is 11 the output is 11 Ex: If the input is 9 or any number not between 11 and 100 (inclusive), the output is: Input must be 11-100 For coding simplicity, follow each output number by a space, even the last one. Use a while loop. Compare the digits do not write a large if-else for all possible same-digit numbers (11,22,33...99), as that approach would be cumbersome for larger ranges. 29999.1122120 LAB ACTIVITY 4.13.1: LAB: Countdown until matching digits 7/10 main. c Load default template... 1 include 2 3 int main() 4 int n, i: 5 scanf("%d", &n); if (n < 20 Il n98) { 7 printf("%d",n); } else { i-n: 1e while (1) printf("%d", 1); if (icie - i/10) 13 14 15 16 17 } 18 printf("\n"); 19 return : 20 ) 12 break; Latest submission - 11:16 PM on 02/05/21 Total score: 7/10 Only show failing tests Download this submission 1: Compare output A 3/3 Input Your output 93 92 91 90 89 88 2: Compare Output A Input 11 Your output 11 3: Compare output a 3/3 Input 20 Your output 20 19 18 17 16 15 14 13 12 11 4: Compare output A 0/2 Output differs. See highlights below. Special character legend Input 101 Your output 101 Expected output Input must be 11-100 5. Compare output A 0/1 Output differs. See highlights below. Special character legend Input Your output Expected output Input must be 11-100

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Answer from: brooke2337

The program in C is as follows:

#include <stdio.h>

int main(){

int n;

scanf("%d", &n);

if(n<11 || n>100){

printf("Input must be between 11 - 100"); }

else{

while(n%11 != 0){

printf("%d ", n);

n--; }

printf("%d ", n); }

return 0;

}

Explanation:

The existing template can not be used. So, I design the program from scratch.

This declares the number n, as integer

int n;

This gets input for n

scanf("%d", &n);

This checks if input is within range (11 - 100, inclusive)

if(n<11 || n>100){

The following prompt is printed if the number is out of range

printf("Input must be between 11 - 100"); }

else{

If input is within range, the countdown is printed

while(n%11 != 0){ This checks if the digits are identical using modulus 11

Print the digit

printf("%d ", n);

Decrement by 1

n--; }

This prints the last expected output

printf("%d ", n); }

Answer from: Quest

My best guess would be b.hope this !

Answer from: Quest

You should always remember that sql is pretty useless to learn about since most sites aren't vulnrable to sql anymore. thedatabase management of an sql database is done with php the actual database is a sql database, but websites interact with databases using php. php coukd mean the site is vulnerable to sqli, but that is only an issue if proper measures aren't put into place, it can be. you don't need php in the url for it to be php it's the code that runs the actual site,it manages databases, actions etc. php runs the whole backend of a site. you could try to put an apostrophe in the url or password input to see if there is an error. if the domain name isnt vulnerable to sqli, that means every file path of the domain probably isn't vulnerable because they are aware of sqli. you should also see what fourm script the site is running. for example, if they are running mybb, there could be a vuln in mybb itself or one of the plugins it uses. another thing you should be able to do is identify if a site even uses sql. for example, a site like roblox may use sql because it has accounts and inorder to have accounts php is used. another way would be to look and see if the site has for example id=1 add a ' after it. if the site is vulnerable you'll see a sql error. but no one cares about sqli it is pretty much dead so telling you this is pointless.

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