hdu5175---Misaki's Kiss again

Problem Description
After the Ferries Wheel, many friends hope to receive the Misaki’s kiss again,so Misaki numbers them 1,2…N−1,N,if someone’s number is M and satisfied the GCD(N,M) equals to N XOR M,he will be kissed again.

Please help Misaki to find all M(1<=M<=N).

Note that:
GCD(a,b) means the greatest common divisor of a and b.
A XOR B means A exclusive or B

Input
There are multiple test cases.

For each testcase, contains a integets N(0

/*************************************************************************    > File Name: bc30-b.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年02月14日 星期六 19时32分44秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double pi = acos(-1);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;vector <LL> yueshu, ans;LL gcd (LL A, LL B){    return B ? gcd (B, A % B) : A;}int main (){    LL n;    int icase = 1;    while (~scanf("%I64d", &n))    {        yueshu.clear();        int tmp = sqrt(n + 0.5);        for (int i = 1; i <= tmp; ++i)        {            if (n % i == 0)            {                yueshu.push_back (i);                if (i != n / i)                {                    yueshu.push_back (n / i);                }            }        }        ans.clear();        int size = yueshu.size();        for (int i = 0; i < size; ++i)        {            LL u = yueshu[i];            if ((u ^ n) <= n && (u ^ n) != 0 && gcd (n, u ^ n) == u)            {                ans.push_back (u ^ n);            }        }        printf("Case #%d:\n", icase++);        size = ans.size();        sort (ans.begin(), ans.end());        printf("%d\n", size);        if (size != 0)        {            printf("%I64d", ans[0]);            for (int i = 1; i < size; ++i)            {                if (ans[i] == ans[i - 1])                {                    continue;                }                printf(" %I64d", ans[i]);            }        }        printf("\n");    }    return 0;}