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

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

 

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.

   Sample test(s) 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).

   
   

    

    

   求出一個最長的序列，使得序列中存在一個所有數的公約數

    

   #include 
   #include 
   #include 
   #include 
   #include 
   #define N 300009
   
   using namespace std;
   
   int a[N];
   int n;
   int ans[N];
   
   int main()
   {
       while(~scanf("%d",&n))
       {
           for(int i=1;i<=n;i++)
           scanf("%d",&a[i]);
   
           int le,ri,tmp=-1,num=0;
   
           for(int i=1;i<=n;)
           {
               le=ri=i;
   
               while(le && a[le]%a[i]==0) le--;
   
               while(ri<=n && a[ri]%a[i]==0) ri++;
   
               i=ri;
   
               ri=ri-le-2;
   
               if(ri>tmp)
               {
                   num=0;
                   tmp=ri;
               }
   
               if(ri==tmp)
               ans[num++]=le+1;
           }
   
           printf("%d %d\n",num,tmp);
   
           for(int i=0;i