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。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<queue>
 7 #include<stdlib.h>
 8 #include<set>
 9 #include<vector>
10 using namespace std;
11 typedef long long LL;
12 LL ans[100000];
13 LL gcd(LL n,LL m);
14 int main(void)
15 {
16     LL N;
17     int cn = 0;
18     while(scanf("%lld",&N)!=EOF)
19     {
20         cn++;
21         ans[0] = -1;
22         printf("Case #%d:\n",cn);
23         if(N == 1)
24         {
25             printf("0\n");
26             printf("\n");
27         }
28         else
29         {
30             int sum = 0;
31             LL i;
32             for(i = 1; i <= sqrt(N); i++)
33             {
34                 if(N%i==0)
35                 {
36                     LL pp = gcd(N,(LL)(N/(LL)i-(LL)1)*(LL)(i));
37                     LL ac = N^((LL)(N/(LL)i-(LL)1)*(LL)(i));
38                     if((N/i-1>=1)&&pp == ac)
39                     {
40                         ans[sum++] = (LL)(N/(LL)i-(LL)1)*(LL)(i);
41                     }
42                     if(N/(LL)i!=i)
43                     {
44                         LL pp = gcd(N,(LL)(i-1)*(LL)(N/i));
45                         LL ac = (LL)(i-1)*(LL)(N/i)^N;
46                                 if(i-1>0&&ac == pp)
47                         {
48                             //printf("1\n");
49                             ans[sum++] = (LL)(i-1)*(LL)(N/i);
50                         }
51                     }
52                 }
53             }
54             printf("%d\n",sum);
55             sort(ans,ans+sum);
56             if(sum>=1)
57                 printf("%lld",ans[0]);
58             for(i = 1; i < sum; i++)
59             {
60                 printf(" %lld",ans[i]);
61             }
62             printf("\n");
63         }
64     }
65     return 0;
66 }
67 LL gcd(LL n,LL m)
68 {
69     if(m == 0)
70         return n;
71     else return gcd(m,n%m);
72 }