Student Locker puzzle

/* 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;

if I substitute "if(locker[i-1] == 1)" with "else" why the program doesn't work? Correct result is opened 1 closed 99. If I use 'else' result becomes opened 50 and closed 50

}
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.

Solution

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,...