题目大意:
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
解题思路:
gcd(N,M) == N XOR M ,则 N XOR M 是 N的一个约数,枚举N的所有约数,然后求GCD判断。
#include
#include
#include
#include
#include
#include
#include
#define LL long long
using namespace std;
const int maxn = 10000000 + 10;
LL ans[maxn];
LL gcd(LL a, LL b)
{
return b == 0 ? a : gcd(b, a % b);
}
int main()
{
LL N;
int kcase = 1;
while(scanf("%I64d", &N)!=EOF)
{
LL sum = 0;
LL R = (LL)(sqrt(N + 0.5));
for(LL i=1;i<=R;i++)
{
if(N % i == 0)
{
if((N ^ i) >=1 && (N ^ i) <= N && gcd((N ^ i),N) == i)
ans[++sum] = (N ^ i);
if(i != N / i)
{
LL x = N / i;
if((N ^ x)>=1 && (N ^ x) <= N && gcd((N ^ x),N)==x)
ans[++sum] = (N ^ x);
}
}
}
printf("Case #%d:\n", kcase++);
printf("%I64d\n", sum);
sort(ans+1,ans+1+sum);
if(sum > 0)
{
printf("%I64d", ans[1]);
for(int i=2;i<=sum;i++) printf(" %I64d", ans[i]);
}
printf("\n");
}
return 0;
}