Make It One CodeForces - 1043F (数论,最短路,好题)

大意: 给定序列$a$, 求最小子集, 使得gcd为1.

 

对于数$x$, 素因子多少次幂是无关紧要的, 这样就可以用一个二进制数来表示.

$x$取$gcd$后的二进制状态最多$2^7$, 可以暴力枚举后继$y$, 可以得到方案数为$sum=\sum\limits_{i=1}^n[gcd(a_i,x)=y]=\sum\limits_{d|\frac{x}{y}}\mu(d)cnt[yd]$.

($cnt[x]$为能被$x$整除的$a_i$个数).

若$sum>0$则可以达到这个后继. 这样跑一次$bfs$即可.

$bfs$的复杂度是A072047的三次幂求和, 打个表发现是N为3e5时只有5412256, 可以通过.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 3e5+10;
int n, mu[N], gpf[N], d[N], cnt[N];
vector<int> fac[N], dv, val;
queue<int> q;

void dfs(int d, int s, int num) {
	val[s] = num;
	if (d!=dv.size()) dfs(d+1,s,num),dfs(d+1,s|1<<d,num*dv[d]);
}

void solve(int x) {
	dv.clear();
	int t = x;
	while (t!=1) { 
		int z = gpf[t];
		while (t%z==0) t/=z;
		dv.pb(z);
	}
	int mx = 1<<dv.size();
	val.resize(mx);
	dfs(0,0,1);
	REP(i,0,mx-1) {
		int y = val[i];
		int sum = 0, r = ~i&(mx-1);
		for (int j=r; j; j=(j-1)&r) {
			sum += mu[val[j]]*cnt[val[j]*y];
		}
		sum += mu[1]*cnt[y];
		if (sum&&!d[y]) {
			q.push(y);
			d[y] = d[x]+1;
		}
	}
}

int main() {
	mu[1] = gpf[1] = 1;
	REP(i,1,N-1) { 
		if (!gpf[i]) for (int j=i;j<N;j+=i) gpf[j]=i;
		for (int j=i;j<N;j+=i) fac[j].pb(i);
		for (int j=2*i;j<N;j+=i) mu[j]-=mu[i];
	}
	scanf("%d", &n);
	REP(i,1,n) {
		int t;
		scanf("%d", &t);
		set<int> s;
		while (t!=1) s.insert(gpf[t]),t/=gpf[t];
		for (int x:s) t*=x;
		if (d[t]) continue;
		d[t] = 1;
		for (int x:fac[t]) ++cnt[x];
		q.push(t);
	}
	if (d[1]) return puts("1"),0;	
	while (q.size()) {
		int x = q.front(); q.pop();
		solve(x);
	}
	printf("%d\n", d[1]?d[1]:-1);
}