HDU 5175 Misaki's Kiss again (异或运算，公式变形)

Misaki's Kiss again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 201    Accepted Submission(s): 57

Problem Description

After the Ferries Wheel, many friends hope to receive the Misaki's kiss again,so Misaki numbers them1,2...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

3515

 

Sample Output

Case #1:12Case #2:14Case #3:310 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

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5175

题目大意：找到满足gcd(N,M)==NxorM的M值(1<=M<=N)

题目分析：令M = N xor K，原式：gcd(N，N xor K) == N xor (N xor K) == K，由此我们可以发现K是N的约数，找到所有N的约数，判断是不是满足那个等式即可，因为是异或运算，结果可能比约数本身大，如1xor2==3，还有异或出来结果等于0的舍掉(它本身)gcd(n,n) != nxorn，还有就是0的时候多输出一个空行，不然pe

#include <cstdio>#include <cmath>#define ll long longint const MAX = 1e5;ll fac[MAX], ans[MAX];ll gcd(ll a, ll b){    return b ? gcd(b, a % b) : a;  }int main(){    int ca = 1;    ll n;    while(scanf("%I64d", &n) != EOF)    {        int cnt1 = 0, cnt2 = 0;        ll tmp = sqrt(n);        for(ll i = 1; i <= tmp; i++)            if(n % i == 0)                fac[cnt1++] = i;        for(ll i = tmp; i >= 1; i--)            if(n % i == 0 && i != 1)                fac[cnt1++] = n / i;        for(int i = cnt1 - 1; i >= 0; i--)            if(fac[i] == gcd(n, n^fac[i]) && (n^fac[i]) != 0 && (n^fac[i]) <= n)                ans[cnt2++] = (n^fac[i]);        printf("Case #%d:\n%d\n", ca++, cnt2);        if(cnt2 == 0)            printf("\n");        for(int i = 0; i < cnt2; i++)        {            if(i != cnt2 - 1)                printf("%I64d ", ans[i]);            else                printf("%I64d\n", ans[i]);        }    }   }