BZOJ 1406 密码箱(数论)

很简洁的题目。求出x^2%n=1的所有x<=n的值。 n<=2e9.

直接枚举x一定是超时的。 看看能不能化成有性质的式子。

有 (x+1)(x-1)%n==0，设n=a*b,那么一定有x+1=k1a,x-1=k2b. 不妨设a<=b.那么就能O(sqrt(n))枚举a。

然后再枚举x，验证x是否满足这两个式子。注意不能令x=k1a-1.由于a比较小，枚举x=k2b+1,k2b-1即可。

另外set很好用啊。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

set<LL>::iterator it;
set<LL>S;

int main ()
{
    LL a, b, x, n;
    scanf("%lld",&n);
    for (int i=; i*i<=n; ++i) {
        if (n%i) continue;
        a=i; b=n/i;
        for (int k=; (x=b*k+)<n; ++k) if ((x+)%a==) S.insert(x);
        for (int k=; (x=b*k-)<n; ++k) if ((x-)%a==) S.insert(x);
    }
    for (it=S.begin(); it!=S.end(); ++it) printf("%lld\n",*it);
    return ;
}