CodeForces - 359D Pair of Numbers （单调栈）

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概述

题目链接：http://codeforces.com/problemset/problem/359/D点击打开链接

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.

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).

#include <iostream>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
#include <limits.h>
#include <math.h>
#include <stack>
#include <queue>
using namespace std;
int gcd(int x,int y)
{
    if(x%y==0||y%x==0)
        return 1;
    return 0;
}
int a[300010];
int la[300010];
int ra[300010];
set<int > q;
set<int > ::iterator it;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    stack<int > s;
    for(int i=1;i<=n;i++)
    {
        la[i]=i;
        if(s.empty())
            {
                s.push(i);
            }
        else
        {
            while(!s.empty())
            {
                if(a[s.top()]%a[i]==0)
                {
                    la[i]=la[s.top()];
                    s.pop();
                }
                else if(a[i]%a[s.top()]==0)
                {
                    s.push(i);
                    break;
                }
                else
                {
                    s.push(i);
                    break;
                }
            }
            if(s.empty())
                s.push(i);
        }
    }
    while(!s.empty())
        s.pop();

    for(int i=n;i>=1;i--)
    {
        ra[i]=i;
        if(s.empty())
            {
                s.push(i);
            }
        else
        {
            while(!s.empty())
            {
                if(a[s.top()]%a[i]==0)
                {
                    ra[i]=ra[s.top()];
                    s.pop();
                }
                else if(a[i]%a[s.top()]==0)
                {
                    s.push(i);
                    break;
                }
                else
                {
                    s.push(i);
                    break;
                }
            }
            if(s.empty())
                s.push(i);
        }
    }

    int maxn=0;int cnt=0;

    for(int i=1;i<=n;i++)
        maxn=max(maxn,ra[i]-la[i]);
    for(int i=1;i<=n;)
        if((ra[i]-la[i])==maxn)
            cnt++,q.insert(la[i]),i=(ra[i]+1);
        else
            i++;
    printf("%d %dn",cnt,maxn);
    for(it=q.begin();it!=q.end();it++)
        printf("%d ",(*it));
}