【BestCoder】#Valentine's Day Round

#2 解方程

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 284    Accepted Submission(s): 78

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<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
   
   
   
    
  
  

 

Source

Valentine's Day Round

gcd(N,x) = N ^ x;
gcd(N,x) 的结果肯定是N的约数。不妨设为k   那么 k = N^x;  k和x都是已知。 。。。  然后解方程两边   k^N = N^x^N = x;
所以x = N ^ (N的约数)  这个是x的必要条件。。。
因为N的约数不一定是gcd(N,x);

注意点：long long
              重复




#include <cstring>
#include <stdio.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long LL;
LL n;

LL gcd(LL x,LL y)
{
    if(y == 0)
        return x;
    return gcd(y,x % y);
}

LL arr[100010];
LL arrnums;
LL nums;
int main()
{
    LL Case = 1;
        freopen("1.txt","r",stdin);
    while(scanf("%I64d",&n) != EOF)
    {
        memset(arr,0,sizeof(arr));
        arrnums = nums = 0;
        LL sq = sqrt((double) n);
        for(LL i=1; i<=sq; i++)
        {
            if(n % i == 0)
            {
                LL x = (i^n);
               // if(x == 0) continue;
                if(1 <= x && x <= n && (n ^ x) == gcd(n,x))
                {
                    arr[arrnums++] = x;
                }

                LL j = n / i;

                x = (j ^ n);
               // if(x == 0) continue;
                if(1 <= x && x <= n &&(n ^ x) == gcd(n,x))
                {
                    arr[arrnums++] = x;
                }
            }


        }

        sort(arr,arr+arrnums);


        for(LL i=0; i<arrnums; i++)
        {
            if(i == 0 || arr[i] != arr[i-1])
            {
                arr[nums++] = arr[i];
            }
        }

        printf("Case #%I64d:\n",Case++);
        printf("%I64d\n",nums);

            for(int i=0; i<nums; i++)
            {
                printf("%I64d",arr[i]);
                if(i != nums -1) printf(" ");
            }

            printf("\n");



    }
}