Induction algorithm--coin flipping problem __ algorithm

Q: There are any n coins facing up, each flipping n coins know to flip all n coins to the opposite side. (* indicates positive, 0 means reverse)

The source code is as follows:

/* * * Flip coin problem, set a total of n coins, once allowed to flip n<n, need to flip x step, each coin flipped y times, * * n*y==n*x, that is, n/n==y/x, because to flip, y must be odd, Judge n/n if the simplest score, such as N is not an odd number can not flip success * *
If N is an odd number, then x takes the minimum value for n=y,n=x, if n/n is not the simplest fraction, the simplest fraction n1/n1, according to the above steps * * To determine, if the N1 is odd can be flipped.
*/#include <stdio.h> #include <string.h> char s[100];
int GCB (int a,int b);
int Is_even (int a);


 int turn_over (int n,int n,int x);
      int main () {int n,n,x,y,g;
      Int J;
      printf ("Please enter the number of coins: \ n");
      scanf ("%d", &n);
          Do {printf ("Please enter the number of coins to flip once: \ n");
          scanf ("%d", &n);
          g = GCB (N,N);
          y = n/g; if (Is_even (y) ==0) printf ("cannot be flipped successfully.")
      \ n ");

      }while (Is_even (y) ==0);
      x = n/g;
      printf ("Initial state: \ n");
          for (j=0;j<n;j++) {s[j]= ' * ';
      printf ("%3c", ' * ');
      } printf ("\ n");

       Turn_over (N,N,X);
  return 0;
    } int GCB (int a,int b)//Two number of greatest common divisor {int c;
        if (b>a) {c = A;
        A = b;
    b = C;
} while (b! = 0) {        c = a%b;
        A = b;
    b = C;
} return A;
    } int Is_even (int a)//to determine if an even number is, then return 0 {while (a > 0) {a = A-2;
} return A;
    } int turn_over (int n,int n,int x)/* Implements the rollover function */{int i,j,k;  for (i = 0;i<x;i++) {j = n/x* (i);
      The starting position of the i+1 step rollover printf ("Step%d: \ n", i+1);
      Getch ();
          for (k=1;k<=n;k++)//k count function, which is to flip n {if (s[j]== ' 0 ') s[j] = ' * ' At each start from s[j];
          else s[j] = ' 0 ';  j = (j+1)%N;
          J reaches the upper limit, but does not flip enough N to flip from the beginning} for (j = 0;j<n;j++) printf ("%3c", S[j]);
    printf ("\ n");
} return 0;
 }

Is relatively simple, there should be no doubt