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 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

 
  就是找出现最多次数的那个数的首尾编号，若是有次数相同的且是最大的，那就

要首尾编号相减最少的那个首尾。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
struct S
{
int cnt;
int Strat,End;
int sum;
S()
{
cnt=0;sum=0;
}
friend bool operator<(const S& a,const S& b)
{
return a.cnt>b.cnt||(a.cnt==b.cnt&&a.sum<b.sum)||(a.cnt==b.cnt&&a.sum==b.sum&&a.Strat<b.Strat);
}
};
S cnt[1000001],ct[100001];
int a[100001];
int main()
{
int n,m,i,j;
scanf("%d",&n);
for(i=1,j=0;i<=n;++i)
{
scanf("%d",&m);
if(cnt[m].cnt)
{
cnt[m].End=i;
}
else
{
a[j++]=m;
cnt[m].Strat=i;
cnt[m].End=i;
}
cnt[m].cnt++;
}
for(i=0;i<j;++i)
{
ct[i]=cnt[a[i]];
ct[i].sum=ct[i].End-ct[i].Strat;
}
sort(ct,ct+j);
/*for(i=0;i<j;++i)
{
printf("%d\n",ct[i].sum);
}*/
printf("%d %d\n",ct[0].Strat,ct[0].End);
return 0;
}