Codeforces Round #312 (Div. 2) B. 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

这道题对技巧和思维都是不错的锻炼。求出现最多次数的那个数在保持出现次数不变的情况下数组的最短长度。

思路：

1用一个vis数组来标记数组中每个数字出现的次数，可以边输入边判断，在线性时间内处理处出现次数最多的次数。注意不是那个数，因为出现的最大次数相同的数可能有多个。

2线性时间扫一遍，处理处每个数的最左和最右端点的编号和区间的长度。

3线性时间再扫一遍，这次只要在出现最大次数相同的数之间比较区间，即可找出出现次数最大且长度最短的子数组。

<pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
int a[maxn];
int vis[maxn];
bool flag[maxn];
int l[maxn];
int r[maxn];
int sum[maxn];
int main()
{
int n;
//freopen("in.txt","r",stdin);
while(~scanf("%d",&n)){
int mas=0;
for(int i=0;i<maxn;i++){
vis[i]=0;sum[i]=0;flag[i]=0;
}
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
vis[a[i]]++;
mas=max(mas,vis[a[i]]);
}
for(int i=1;i<=n;i++){
if(!flag[a[i]]){
flag[a[i]]=1;
l[a[i]]=i;
}
r[a[i]]=i;
sum[a[i]]=r[a[i]]-l[a[i]]+1;
}
int ll=0,rr=0;
int longest=inf;
for(int i=1;i<=n;i++){
if(vis[a[i]]==mas){
if(sum[a[i]]<longest){
longest=sum[a[i]];
ll=l[a[i]];
rr=r[a[i]];
}
}
}
printf("%d %d\n",ll,rr);
}
}