#CF 359 D Pair of Numbers（dp）

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本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR增强现实、SLAM、物体检测与识别、语音识别与变声等。探讨了这些技术在实际应用中的作用与价值。

Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r(1 ≤ l ≤ r ≤ n), such that the following conditions hold:

there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
value r - l takes the maximum value among all pairs for which condition 1 is true;

Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.

Input

The first line contains integer n (1 ≤ n ≤ 3·105).

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).

Output

Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.

Sample test(s)

input
5
4 6 9 3 6

output
1 3
2 

input
5
1 3 5 7 9

output
1 4
1 

input
5
2 3 5 7 11

output
5 0
1 2 3 4 5 

Note

In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.

In the second sample all numbers are divisible by number 1.

In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).

恶心的题：需要注意去重

题意：一个区间l,r,纯在一个x (l<=x<=r)使得整个区间的数都可以整除a[x]),求r-l最大的情况，如有多组，按l小的输出

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define pii pair<int,int>

#define eps 1e-8


#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f

#define N 1000000

vector<pii>ans;

int a[N],le[N],ri[N];
int vis[N];

int n;

int main()
{
	int i,j;
	while(~sf(n))
	{
		fre(i,1,n+1)
		  sf(a[i]);

		for(i=1;i<=n;i++)
		{
			le[i]=i;
			while(le[i]>1&&a[le[i]-1]%a[i]==0)
				le[i]=le[le[i]-1];
		}

		for(i=n;i>=1;i--)
		{
			ri[i]=i;
			while(ri[i]<n&&a[ri[i]+1]%a[i]==0)
				ri[i]=ri[ri[i]+1];
		}

        ans.clear();
        for(i=1;i<=n;i++)
		{
			int temp=ri[i]-le[i];
			if(ans.size()==0)
			{
				ans.push_back(make_pair(le[i],ri[i]-le[i]));
				continue;
			}
            if(temp>ans[0].second)
			{
				ans.clear();
				ans.push_back(make_pair(le[i],ri[i]-le[i]));
			    continue;
			}
			if(temp==ans[0].second)
			{
			  ans.push_back(make_pair(le[i],ri[i]-le[i]));
			}
		}

		int all=0;
		int pre=-1;
		for(i=0;i<ans.size();i++)
			if(ans[i].first!=pre)
		{
			all++;
			pre=ans[i].first;
		}
		pf("%d %d\n",all,ans[0].second);
        pre=-1;
        int i=0;

		for(i=0;i<ans.size();i++)
		 {
            if(ans[i].first==pre) continue;    //防止有多个数相同，那么区间都是一个，需要去重
		 	pre=ans[i].first;
		 	if(i) pf(" ");
		 	pf("%d",pre);
		 }

      pf("\n");
	}
  return 0;
}