troublefish - 1 year ago 88

C Question

/* There are 100 students and 100 lockers. Student 1 opens all, student 2 closes every second one, student 3 changes every third

locker(closes if open, opens if close), Student 4 changes every forth locker and so on for all 100 students.

Which locker will be left open? */

Here is my code thus far:

`#include <stdio.h>`

int main(void)

{

int locker[100], i, closed = 0, opened = 0;

for(i=0; i<100; i++) locker[i] = 1;//0 means closed locker, 1 means open locker

for(i=1; i<101; i++) if(i % 2 == 0) locker[i-1] = 0; // every second locker is closed by second student...(2,4,6,7)

for(i=3; i<101; i++){ // i means student no. i

if(locker[i-1] == 0) locker[i-1] = 1;

if(locker[i-1] == 1) locker[i-1] = 0;

`}`

for(i=0; i<100; i++){

if(locker[i] == 0) closed = closed + 1;

else opened = opened + 1;

}

printf("opened locker %d\nclosed locker %d", opened, closed);

return 0;

}

This is my first post in stack overflow. Correct me if I've done anything wrong.

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Answer Source

I'll give you a few hints to help you out.

- The answer is that 10 lockers remain open, 90 are closed.
- For this particular problem, it's easier to write the code if you avoid zero-based indexing. So
declare the array as
`int locker[101];`

and then use indexes 1 thru 100 to represent the 100 lockers. - The Nth student is supposed to change every Nth locker. So you need
two nested
`for`

loops. The outer loop keeps track of`n`

, and the inner loop flips lockers. The inner loop that only affects every Nth locker should look like this

`for ( i = n; i <= 100; i += n ) // every Nth locker locker[i] = 1 - locker[i]; // flip the locker`

Note that instead of the normal

`i=0`

and`i++`

, we have`i=n`

and`i+=n`

. So, for example, if`n`

is 3, then the values of`i`

are`3,6,9,...`