int main()
{
int n;
sieve();
while (scanf("%d", &n) == 1 && n != 0)
printf("%d\n", survivor(n));
return 0;
}
int survivor(int n, int flag, int j, int d)
{
int i;
if(flag == 1)
d += pow(2, j-1);
else if(flag == 2)
d -= pow(2, j-1);
if(n==1)
{
return (int)(pow(2, j) - d);
}
if(flag == 0)
{
for(i = 1; i <= n/2; i++)
printf("%d ", i*2);
if(n%2 !=0)
printf("1 ");
j = 1;
if(n%2 == 0)
return survivor(n/2, 1, j, d);
else
return survivor(n/2, 2, j, d);
}
for(i = 1; i <= n/2; i++)
{
printf("%d ", (int)(pow(2, j)*2*i - d));
}
if(n%2 != 0)
{
printf("%d ", (int)(pow(2, j)*1 - d));
j++;
return survivor(n/2, 2, j, d);
}
else
{
j++;
return survivor(n/2, 1, j, d);
}
}
int main()
{
int n, v, j = 0, d = 0, m, l;
printf("\nEnter n : ");
scanf("%d", &n);
printf("The elimination order : ");
v = survivor(n, 0, j, d);
printf("\nThe survivor is : %d\n", v);
m = log(n)/log(2);
l = n - pow(2, m);
v = 2*l + 1;
printf("\nUsing the formula V(2^m + l) = 2*l + 1, the survivor is: %d\n", v);
return 0;
}
int main(void)
{
assert (survivor(5) == 3);
assert (survivor(10) == 5);
assert (survivor(42) == 21);
assert (survivor(66000000) == 64891137);
assert (decode("05e0") == 5);
assert (decode("01e1") == 10);
assert (decode("42e0") == 42);
assert (decode("66e6") == 66000000);
char s[5];
while (1)
{
scanf("%s\n", s);
if (strcmp(s, "00e0") == 0) break;
printf("%ld\n", survivor(decode(s)));
}
return 0;
}
long int survivor(const long int n)
{
if (n <= 2) return 1;
else if (n%2 == 0) return 2*survivor(n/2) - 1;
else return 2*survivor((n-1)/2) + 1;
}
