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:

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

const int M = 1e6+;

int n,m,k,tot=,tim=;

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]=;

    int ret=;

    int u=b[p];

    int ll,rr;

    ++tot;

    for(int i=; i<=n; i++) {

        if(!l[tot]) {

            ll=p-i;

            if(ll>=&&b[ll]%u==) {

                vis[ll]=;

            } else l[tot]=ll+,ret++;

        }

        if(!r[tot]) {

            rr=p+i;

            if(rr<=n&&b[rr]%u==) {

                vis[rr]=;

            } else r[tot]=rr-,ret++;

        }

        if(ret==)return;

    }

}

int main() {

    scanf("%d",&n);

    for(int i=; 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=;i<n;i++){

        int p=vec[i].second;

        if(!vis[p])Find(p);

    }

    for(int i=; i<=tot; i++) {

        ans[i-].l=l[i];

        ans[i-].r=r[i];

        ans[i-].len=r[i]-l[i];

    }

    sort(ans,ans+tot,cmp);

    int t=ans[].len,ret=;

    for(int i=;i<tot;i++){

        if(ans[i].len<t)break;

        ret++;

    }

    printf("%d %d\n",ret,t);

    for(int i=;i<ret;i++){

        Ans.pb(ans[i].l);

    }

    sort(Ans.begin(),Ans.end());

    for(int i=;i<ret;i++){

        printf("%d ",Ans[i]);

    }

    printf("\n");

    return ;

}

#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+;

const int M = 1e6+;

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=; i<=n; ++i)scanf("%d",&a[i]),ans_l[i]=ans_r[i]=i;

}

void work() {

    for(int i=; i<=n; ++i)for(; ans_l[i]>&&a[ans_l[i]-]%a[i]==;)ans_l[i]=ans_l[ans_l[i]-];

    for(int i=n; i>=; --i)for(; ans_r[i]<n&&a[ans_r[i]+]%a[i]==;)ans_r[i]=ans_r[ans_r[i]+];

    for(int i=; 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=]=L;

    }

}

void outit() {

    sort(ans+,ans+Ans+);

    for(int i=; i<=Ans; ++i)if(ans[i]!=ans[i-])++xllend3;

    printf("%d %d\n%d",xllend3,len,ans[]);

    for(int i=; i<=Ans; ++i)if(ans[i]!=ans[i-])printf(" %d",ans[i]);

    puts("");

}

int main() {

    init();

    work();

    outit();

    return ;

}

巴特西

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

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