数学(hdu5175Misaki's Kiss again)

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 820    Accepted Submission(s): 205


Problem Description
After the Ferries Wheel, many friends hope to receive the Misaki's kiss again,so Misaki numbers them 1,2...N1,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<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
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


思路:暴力枚举所有的因子,在判断是不是符合条件

注意,有可能异火的结果大于N,要判断一下

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=101000;
typedef long long LL;
LL fac[maxn],ans[maxn];
LL N;
int cnt;
int cas;
void init()
{
    LL tmp=sqrt(N);
    cnt=0;
    for(LL i=1;i<=tmp;i++)
        if(N%i==0)
        {
            fac[cnt++]=i;
            if(i*i!=N){fac[cnt++]=N/i;}
        }
}
LL gcd(LL a,LL b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
void solve()
{
    int num=0;
    for(int i=0;i





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