Codeforces 558B Amr and The Large Array

Source: Internet
Author: User
Tags cmath

Codeforces 558B Amr and The Large Array
B. Amr and The Large Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Amr has got a large array of sizeN. Amr doesn't like large arrays so he intends to make it smaller.

Amr doesn't care about anything in the array doesn't the beauty of it. the beauty of the array is defined to be the maximum number of times that some number occurs in this array. he wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.

Help Amr by choosing the smallest subsegment possible.

Input

The first line contains one numberN(1 digit ≤ DigitNLimit ≤ limit 105), the size of the array.

The second line containsNIntegersAI(1 digit ≤ DigitAILimit ≤ limit 106), representing elements of the array.

Output

Output two integersL, Bytes,R(1 digit ≤ DigitLLimit ≤ limitRLimit ≤ limitN), The beginning and the end of the subsegment chosen respectively.

If there are several possible answers you may output any of them.

Sample test (s) input
51 1 2 2 1
Output
1 5
Input
51 2 2 3 1
Output
2 3
Input
61 2 2 1 1 2
Output
1 5
Note

A subsegmentBOf an arrayAFromLToRIs an array of sizeRAccept-Encoding-LLimit + limit 1 whereBISignature = SignatureALRegion + RegionIAccept-limit 1 for all 1 records ≤ limitILimit ≤ limitRAccept-Encoding-LLimit + limit 1

 

 

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        using namespace std;const double PI = acos(-1.0);const double e = 2.718281828459;const double eps = 1e-8;struct node{ int num; int time; int l; int r; int range;} index[1000010];int main(){ //freopen(in.txt, r, stdin); //freopen(out.txt, w, stdout); int n; while(cin>>n) { map
         
          g; g.clear(); memset(index, 0, sizeof(index)); int t; int d = 0; int dd; int maxd = 1; int maxtime = 0; int maxrange = 1; for(int i = 1; i <= n; i++) { scanf(%d, &t); if(g[t] != 0) { dd = g[t]; index[dd].time++; index[dd].r = i; index[dd].range = index[dd].r-index[dd].l+1; if(maxtime
          
           index[dd].range) { maxtime = index[dd].time; maxd = dd; maxrange = index[dd].range; } } else { g[t] = ++d; dd = g[t]; index[dd].num = t; index[dd].time = 1; index[dd].l = index[dd].r = i; index[dd].range = 1; if(maxtime < index[dd].time) { maxtime = index[dd].time; maxd = dd; maxrange = 1; } } } printf(%d %d, index[maxd].l, index[maxd].r); } return 0;}
          
         
       
      
     
    
   
  
 


 

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