CodeForces - 359D D. Pair of Numbers

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, .
4000
.., 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 300005

using namespace std;

int a
,ans
;

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 300005

using namespace std;

int a
,ans
,ll
,rr
;
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;
}