CodeForces

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:

1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
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, ..., 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

54 6 9 3 6

output

1 32 

input

51 3 5 7 9

output

1 41 

input

52 3 5 7 11

output

5 01 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).

题意： n个数 找出一个尽量长的区间l和r 使得l和r中的一个数可以整除区间里的所有数

输出：

区间的个数（如果最长区间有多个则输出多个） 区间的长度-1

区间的左边界

思路：枚举每一个数的左边界和右边界，但是根据数据范围可以得出不可以用n方的算法，会超时，所以在处理左右边界可以优化一下。其他的就比较简单了。下面贴上两种代码

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 300005using namespace std;int a[N],ans[N];int main(){int i,l,r,n;scanf("%d",&n);for(i=1;i<=n;i++) scanf("%d",&a[i]);int cnt=0;int len=0;for(i=1;i<=n;){l=i;r=i;while(l>=1&&a[l]%a[i]==0) l--;while(r<=n&&a[r]%a[i]==0) r++;l++;r--;//printf("left : %d right : %d\n",l,r);i=r+1;int num=r-l;if(num==len) ans[++cnt]=l;else if(num>len){cnt=0;len=num;ans[++cnt]=l;}}printf("%d %d\n",cnt,len);for(i=1;i<=cnt;i++){if(i==cnt) printf("%d\n",ans[i]);else printf("%d ",ans[i]);}return 0;}

第二种最后输出要注意一下.

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 300005using namespace std;int a[N],ans[N],ll[N],rr[N];int n;void init(){int i;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);ll[i]=rr[i]=i;}return ;}void output(){int i,j,left,right;int Ans=0,len=0;for(i=1;i<=n;++i)    {        left=ll[i];right=rr[i];        if(right-left==len)ans[++Ans]=left;        if(right-left>len)len=right-left,ans[Ans=1]=left;    }    int xllend3=0;    sort(ans+1,ans+Ans+1);    for(i=1;i<=Ans;++i)if(ans[i]!=ans[i-1])++xllend3;    printf("%d %d\n%d",xllend3,len,ans[1]);    for(i=2;i<=Ans;++i)if(ans[i]!=ans[i-1])printf(" %d",ans[i]);    printf("\n");}int main(){int i,j,l,r;init();int maxx=0;int cnt=0;for(int i=1;i<=n;++i)for(;ll[i]>1&&a[ll[i]-1]%a[i]==0;)ll[i]=ll[ll[i]-1];    for(int i=n;i>=1;--i)for(;rr[i]<n&&a[rr[i]+1]%a[i]==0;)rr[i]=rr[rr[i]+1];output();return 0;}