description
analysis
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这题出的失败在只卡正解不卡暴力
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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
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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)))
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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
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时间复杂度(O(nlog_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
",cnt,mx);
fo(i,1,cnt)printf("%lld ",g[i]);
return 0;
}