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 ai (1 ≤ ai ≤ 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 Bi = Al + i - 1 for
all 1 ≤ i ≤ r - l + 1

解题说明：此题要求找出重复出现次数最多的数字，并且标记出该数字最前最后出现的位置。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>

using namespace std;
long long int a[1000005],b[1000005];
int main()
{
long long int i,t,l,r,n,max=-1,min=100000000,k;
scanf("%lld",&n);
for(i=1;i<=n;i++)
{
scanf("%lld",&t);
a[t]++;
if(a[t]==1)
{
b[t]=i;
}
if(a[t]>max)
{
max=a[t];
min=i-b[t];
l=b[t];
r=i;
}
else if(a[t]==max)
{
if((i-b[t])<min)
{
max=a[t];
l=b[t];
r=i;
min=r-l;
}
}
}
printf("%lld %lld\n",l,r);
return 0;
}