Misaki‘s Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 513    Accepted Submission(s): 121

Problem Description

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

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

Node 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

sqrt(n) 求出n的所有约数ei。

然后求出 a 符合： a^n == ei  。

再判断 a 跟 n 的 GCD 是否等于ei 即可。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <vector>
#include <queue>

using namespace std ;
typedef long long LL ;
typedef pair<int,int> pii;
#define X first
#define Y second

const int N = 100010;
LL n , e[N] , x[N] ;
int aa[100] , bb[100] , res[100] ;
inline LL gcd( LL a , LL b ) { return b == 0 ? a : gcd( b , a % b );}
LL decode( LL a , LL b ) {
    int lena = 0 , lenb = 0 ;
    memset( aa , 0 , sizeof aa );
    memset( bb , 0 , sizeof bb );
    LL x = a ;
    while( x > 0 ) { aa[lena++] = x % 2 ; x /= 2 ; }
    x = b ;
    while( x > 0 ) { bb[lenb++] = x % 2 ; x /= 2 ; }

    LL ans = 0 ;
    for( int i = 0 ; i < lena ; ++i ) {
        if( bb[i] == 0 ) res[i] = aa[i];
        else res[i] = aa[i]^1;
        if( res[i] ) ans += (1LL<<i);
    }
    return ans ;
}

void solve() {
    int tot = 0 , cnt = 0  ;
    for( LL i = 1 ; i * i <= n ; ++i ) {
        if( n % i == 0 ){
            e[tot++] = i ;
            if( i != n/i ) e[tot++] = n/i ;
        }
    }
    for( int i = 0 ; i < tot ; ++i ) {
       LL tmp = decode( n , e[i] );
       if( tmp < 1 || tmp > n || tmp < e[i] || gcd( n , tmp ) != e[i] ) continue ;
       x[cnt++] = tmp ;
    }
    sort( x , x + cnt );
    printf("%d\n",cnt);
    for( int i = 0 ; i < cnt ; ++i ) {
        printf("%I64d",x[i]);
        if( i < cnt -1 ) printf(" ");
    }
    puts("");
}
int main ()  {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int _ , cas =1 ;
    while( scanf("%I64d",&n) != EOF ) {
        printf("Case #%d:\n",cas++);
        solve();
    }
}

HDU 5175 Misaki's Kiss again