BZOJ2277 [Poi2011]Strongbox 【数论】

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 mid 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
",ans);
	return 0;
}