D. Pair of Numbers----思维题

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, ..., 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.

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).

题目链接：http://codeforces.com/contest/359/problem/D

题目的意思是给定一个数字，求最长的区间，使得该区间内存在一个元素，它能整除该区间内的每一个元素。

思路就是从数组第一个元素开始向两边延伸，如右边延伸到r，下一轮从r+1开始考虑即可。

代码：

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <vector> #define inf 0x3f3f3f3f using namespace std; const int MAXN=3e5; int a[400000]; int main(){ vector<int>v; int n, ans; scanf("%d",&n); bool flag=false; ans=-inf; for(int i=1;i<=n;++i){ scanf("%d",&a[i]); } for(int i=1;i<=n;++i){ if(a[i]==1){ flag = true; printf("1 %d\n1\n",n-1); break; } int j,k; for(j=i;j-1>0&&a[j-1]%a[i]==0;--j); for(k=i;k+1<=n&&a[k+1]%a[i]==0;++k); if(k-j>ans){ ans = k-j; v.clear(); } if(k-j == ans){ v.push_back(j); } i=k; } if(!flag){ printf("%d %d\n",v.size(),ans); for(int i=0;i<v.size()-1;++i){ printf("%d ",v[i]); } printf("%d\n",v[v.size()-1]); } return 0; }