题目链接

BZOJ2277

题解

orz太难了

如果一个数\(x\)是密码，那么所有\((x,n)\)的倍数都是密码
如果两个数\(x,y\)是密码，那么所有\((x,y)\)的倍数都是密码

那么如果最后的密码集合为\(\{x_i\}\)那么一定存在一个\(x_i\)是剩余所有数的\(gcd\)
所以我们只需找最小的\(x | n\)且\(x | a_k\)且\(x \nmid a_i\)

那就找出\((a_k,n)\)的所有质因子，再用\((a_i,a_k,n)\)筛去不合法的即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 1000005,maxm = 10000005,INF = 1000000000;
inline LL read(){LL out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
LL n,k,a[maxn],L,fac[maxn],p[maxn],tot,cnt;
int vis[maxn];
LL gcd(LL a,LL b){return b ? gcd(b,a % b) : a;}
void getfac(){LL x,tmp = L;for (x = 1; x * x <= tmp; x++){if (tmp % x == 0){fac[++tot] = x;if (x * x != tmp) fac[++tot] = tmp / x;}}sort(fac + 1,fac + 1 + tot);
}
void getp(){LL x,tmp = L;for (x = 2; x * x <=tmp; x++)if (tmp % x == 0){p[++cnt] = x;while (tmp % x == 0) tmp /= x;}if (tmp - 1) p[++cnt] = tmp;
}
int main(){n = read(); k = read();REP(i,k) a[i] = read();L = gcd(n,a[k]);getfac(); getp();LL x;for (int i = 1; i < k; i++){x = gcd(a[i],L);vis[lower_bound(fac + 1,fac + 1 + tot,x) - fac] = true;}for (int i = tot - 1; i; i--){if (vis[i]) continue;x = fac[i];for (int j = 1; j <= cnt && x * p[j] <= L; j++){LL y = x * p[j];int pos = lower_bound(fac + 1,fac + 1 + tot,y) - fac;if (fac[pos] == y && vis[pos]){vis[i] = true;break;}}}LL ans = 0;for (int i = 1; i <= tot; i++)if (!vis[i]){ans = n / fac[i]; break;}printf("%lld\n",ans);return 0;
}