Misaki's Kiss again（hdu5175）

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1773    Accepted Submission(s): 459

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

 

B

Input

There are multiple test cases.

For each testcase, contains a integets N(0<N<=1010)

 

Output

For each test case,
first line output Case #X:,
second line output k means the number of friends will get a kiss.
third line contains k number mean the friends' number, sort them in ascending and separated by a space between two numbers

 

Sample Input

3

5

15

Sample Output

Case #1: 1 2

Case #2: 1 4

Case #3: 3 10 12 14

思路：gcd（n，m） = norm --->gcd(n,nork) =  k;应为m 可以表示为nork的形式，所以，我们只要枚举n的因子k然后判断是否符合即可。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#include<queue>
#include<stdlib.h>
#include<set>
#include<vector>
using namespace std;
typedef long long LL;
LL ans[100000];
LL gcd(LL n,LL m);
int main(void)
{
    LL N;
    int cn = 0;
    while(scanf("%lld",&N)!=EOF)
    {
        cn++;
        ans[0] = -1;
        printf("Case #%d:
",cn);
        if(N == 1)
        {
            printf("0
");
            printf("
");
        }
        else
        {
            int sum = 0;
            LL i;
            for(i = 1; i <= sqrt(N); i++)
            {
                if(N%i==0)
                {
                    LL pp = gcd(N,(LL)(N/(LL)i-(LL)1)*(LL)(i));
                    LL ac = N^((LL)(N/(LL)i-(LL)1)*(LL)(i));
                    if((N/i-1>=1)&&pp == ac)
                    {
                        ans[sum++] = (LL)(N/(LL)i-(LL)1)*(LL)(i);
                    }
                    if(N/(LL)i!=i)
                    {
                        LL pp = gcd(N,(LL)(i-1)*(LL)(N/i));
                        LL ac = (LL)(i-1)*(LL)(N/i)^N;
                                if(i-1>0&&ac == pp)
                        {
                            //printf("1
");
                            ans[sum++] = (LL)(i-1)*(LL)(N/i);
                        }
                    }
                }
            }
            printf("%d
",sum);
            sort(ans,ans+sum);
            if(sum>=1)
                printf("%lld",ans[0]);
            for(i = 1; i < sum; i++)
            {
                printf(" %lld",ans[i]);
            }
            printf("
");
        }
    }
    return 0;
}
LL gcd(LL n,LL m)
{
    if(m == 0)
        return n;
    else return gcd(m,n%m);
}