3
5
15

Case #1:
1
2
Case #2:
1
4
Case #3:
3
10 12 14

Hint
In the third sample, gcd(15,10)=5 and (15 xor 10)=5, gcd(15,12)=3 and (15 xor 12)=3,gcd(15,14)=1 and (15 xor 14)=1
 

#include <cstdio>
#include <cmath>
#define ll long long
int const MAX = 1e5;
ll fac[MAX], ans[MAX];

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;  
}

int main()
{
    int ca = 1;
    ll n;
    while(scanf("%I64d", &n) != EOF)
    {
        int cnt1 = 0, cnt2 = 0;
        ll tmp = sqrt(n);
        for(ll i = 1; i <= tmp; i++)
            if(n % i == 0)
                fac[cnt1++] = i;
        for(ll i = tmp; i >= 1; i--)
            if(n % i == 0 && i != 1)
                fac[cnt1++] = n / i;
        for(int i = cnt1 - 1; i >= 0; i--)
            if(fac[i] == gcd(n, n^fac[i]) && (n^fac[i]) != 0 && (n^fac[i]) <= n)
                ans[cnt2++] = (n^fac[i]);
        printf("Case #%d:\n%d\n", ca++, cnt2);
        if(cnt2 == 0)
            printf("\n");
        for(int i = 0; i < cnt2; i++)
        {
            if(i != cnt2 - 1)
                printf("%I64d ", ans[i]);
            else
                printf("%I64d\n", ans[i]);
        }
    }   
}