Codeforces Round #209 (Div. 2) D. Pair of Numbers （模拟）

D. Pair of Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Simon has an array a₁, a₂, ..., a_n, 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:

1. there is integer j (l ≤ j ≤ r), such that all integers a_l, a_l + 1, ..., a_r are divisible by a_j;
2. 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, ..., a_n(1 ≤ a_i ≤ 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.

Examples

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).

【分析】给定一个正整数数组，求最长的区间，使得该区间内存在一个元素，它能整除该区间的每个元素。

 可能我的方法有点复杂吧。我是先从小到大排序，记录所有数的位置，然后从小到大向两边扩展，就行了，详细见代码一。还有种做法类似DP，很短很神奇，见代码二。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
typedef long long ll;
using namespace std;
const int N = 3e5+10;
const int M = 1e6+10;
int n,m,k,tot=0,tim=0;
int head[N],vis[N],l[N],r[N];
int a[N],b[N];
vector<pair<int,int> >vec;
vector<int>Ans;
struct man{
    int l,r,len;
}ans[N];
bool cmp(man s,man d){
    return s.len>d.len;
}
void Find(int p) {
    vis[p]=1;
    int ret=0;
    int u=b[p];
    int ll,rr;
    ++tot;
    for(int i=1; i<=n; i++) {
        if(!l[tot]) {
            ll=p-i;
            if(ll>=1&&b[ll]%u==0) {
                vis[ll]=1;
            } else l[tot]=ll+1,ret++;
        }
        if(!r[tot]) {
            rr=p+i;
            if(rr<=n&&b[rr]%u==0) {
                vis[rr]=1;
            } else r[tot]=rr-1,ret++;
        }
        if(ret==2)return;
    }
}
int main() {
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        scanf("%d",&a[i]);
        b[i]=a[i];
        vec.pb(mp(a[i],i));
    }
    sort(vec.begin(),vec.end());
    for(int i=0;i<n;i++){
        int p=vec[i].second;
        if(!vis[p])Find(p);
    }
    for(int i=1; i<=tot; i++) {
        ans[i-1].l=l[i];
        ans[i-1].r=r[i];
        ans[i-1].len=r[i]-l[i];
    }
    sort(ans,ans+tot,cmp);
    int t=ans[0].len,ret=0;
    for(int i=0;i<tot;i++){
        if(ans[i].len<t)break;
        ret++;
    }
    printf("%d %dn",ret,t);
    for(int i=0;i<ret;i++){
        Ans.pb(ans[i].l);
    }
    sort(Ans.begin(),Ans.end());
    for(int i=0;i<ret;i++){
        printf("%d ",Ans[i]);
    }
    printf("n");
    return 0;
}

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 2e6+10;
const int M = 1e6+10;
int n,L,R,len,Ans,xllend3;
int a[N],ans_l[N],ans_r[N],ans[N];
void init() {
    scanf("%d",&n);
    for(int i=1; i<=n; ++i)scanf("%d",&a[i]),ans_l[i]=ans_r[i]=i;
}
void work() {
    for(int i=1; i<=n; ++i)for(; ans_l[i]>1&&a[ans_l[i]-1]%a[i]==0;)ans_l[i]=ans_l[ans_l[i]-1];
    for(int i=n; i>=1; --i)for(; ans_r[i]<n&&a[ans_r[i]+1]%a[i]==0;)ans_r[i]=ans_r[ans_r[i]+1];
    for(int i=1; i<=n; ++i) {
        L=ans_l[i];
        R=ans_r[i];
        if(R-L==len)ans[++Ans]=L;
        if(R-L>len)len=R-L,ans[Ans=1]=L;
    }
}

void outit() {
    sort(ans+1,ans+Ans+1);
    for(int i=1; i<=Ans; ++i)if(ans[i]!=ans[i-1])++xllend3;
    printf("%d %dn%d",xllend3,len,ans[1]);
    for(int i=2; i<=Ans; ++i)if(ans[i]!=ans[i-1])printf(" %d",ans[i]);
    puts("");
}

int main() {
    init();
    work();
    outit();
    return 0;
}