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Codeforces 558B Amr and The Large Array

編輯：C++入門知識

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 size n. Amr doesn't like large arrays so he intends to make it smaller.

Amr doesn't care about anything in the array except 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 number n (1 ≤ n ≤ 10⁵), the size of the array.

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁶), representing elements of the array.

Output

Output two integers l, r (1 ≤ l ≤ r ≤ n), 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

5
1 1 2 2 1

output

1 5

input

5
1 2 2 3 1

output

2 3

input

6
1 2 2 1 1 2

output

1 5

Note

A subsegment B of an array A from l to r is an array of size r - l + 1 where B_i = A_{l + i - 1} for all 1 ≤ i ≤ r - l + 1

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
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)
    {
        mapg;
        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(maxtimeindex[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;
}