hdu 5175(数论)

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1621    Accepted Submission(s): 414

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

 

题意:找到1=<m<=n里面满足 gcd(n,m) = n xor m 的m的个数.然后输出所有的 m .

题解:数据量 10^10 ，减少到 10^5 就不会超时了.所以我们从异或操作考虑 , n^m = k ---> n^k = m 然后从gcd(n,m)考虑,因为 n<=m 所以 gcd(n,m)必定是 n 的因子,所以我们可以在 O(sqrt(n)) 的时间里面将 n 的因子全部弄出来，然后枚举其因子, n^factor[i] = m ---> gcd(n,m) == factor[i] 那么这个m就是满足条件的,注意一点就是当 m 的个数为 0 的时候，后面要输出空行。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
LL factor[1000000];
LL gcd(LL a,LL b){
    return b==0?a:gcd(b,a%b);
}
LL ans[1000000];
int main()
{
    LL n;
    int t=1;
    while(scanf("%lld",&n)!=EOF){
        factor[1] = 1;
        int id = 2;
        for(LL i=2;i*i<=n;i++){
            if(n%i==0){
                if(i*i==n){
                    factor[id++] = i;
                }else{
                    factor[id++] = i;
                    factor[id++] = n/i;
                }
            }
        }
        factor[id++] = n;
        int cnt = 0;
        for(LL i=1;i<id;i++){
            LL M = n^factor[i];
            if(M<1||M>n) continue;
            if(gcd(n,M)==factor[i]){
                ans[cnt++] = M;
            }
        }
        printf("Case #%d:\n",t++);
        if(cnt==0){
            printf("0\n\n");
        }else{
            sort(ans,ans+cnt);
            printf("%d\n",cnt);
            for(int i=0;i<cnt-1;i++){
                printf("%lld ",ans[i]);
            }
            printf("%lld\n",ans[cnt-1]);
        }
    }
    return 0;
}