【JZOJ6294】动态数点

description

---

analysis

• 这题出的失败在只卡正解不卡暴力

• 比较好想的方法是枚举约数，向两边二分，但是这个不满足二分性

• 首先用\(ST\)表维护区间的\(\gcd\)，不用线段树，这样查询就是\(O(\log_2(\max_{i=1}^{n} a_i))\)

• 然后照上面的方法做就行了，枚举约数，向左右二分，判断区间\(\gcd\)是否等于当前约数

• 时间复杂度\(O(n\log_2n)\)乘\(32\)的常数，注意卡常比如预处理出\(\log\)的值

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[MAXN][20],g[MAXN];
ll a[MAXN],ans[MAXN],LOG[MAXN];
ll n,mx,tot,cnt;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
inline ll query_gcd(ll l,ll r)
{
	ll k=LOG[r-l+1];
	return gcd(f[l][k],f[r+1-(1<<k)][k]);
}
int main()
{
	freopen("T2.in","r",stdin);
	//freopen("point.in","r",stdin);
	//freopen("point.out","w",stdout);
	n=read(),LOG[0]=-1;
	fo(i,1,n)a[i]=f[i][0]=read(),LOG[i]=LOG[i>>1]+1;
	fo(j,1,LOG[n])fo(i,1,n+1-(1<<j))
	f[i][j]=gcd(f[i][j-1],f[i+(1<<(j-1))][j-1]);
	fo(i,1,n)
	{
		ll l=1,r=i,mid,tmp;
		while (l<=r)
		{
			mid=(l+r)>>1;
			query_gcd(mid,i)==a[i]?r=mid-1:l=mid+1;
		}
		tmp=l,l=i,r=n;
		while (l<=r)
		{
			mid=(l+r)>>1;
			query_gcd(i,mid)==a[i]?l=mid+1:r=mid-1;
		}
		if (r-tmp>mx){mx=r-tmp,ans[tot=1]=tmp;}
		else if (r-tmp==mx)ans[++tot]=tmp;
	}
	sort(ans+1,ans+tot+1);ll i=1;
	while (i<=tot)
	{
		g[++cnt]=ans[i];
		while (ans[i]==ans[i+1] && i<=n)++i;
		++i;
	}
	printf("%lld %lld\n",cnt,mx);
	fo(i,1,cnt)printf("%lld ",g[i]);
	return 0;
}