Codeforces

1.题目描述：

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

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 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

2.题意概述：

一个整数序列的beauty值为序列中出现次数最多的整数的出现次数。求与给定整数序列的beauty值相同的该序列的最短子区间，只输出区间的始、末位置。

3.解题思路：

类似桶排序思想，开两个一维数组L和R和T，每读入一个数x，用l记录当前数x为止最小的左区间，r记录当前数x最大的右区间，这样r[i] - l[i] + 1就是可行区间，T记录当前的美丽值，最好扫一遍就是复杂度O(N)

4.代码实现：

#include <cstdio>#include <algorithm>#define INF 0x3f3f3f3f#define maxn 1010000using namespace std;int l[maxn], r[maxn], t[maxn], n;int main(){while (scanf("%d", &n) != EOF){int ma = 0, ans = INF, j = 0;for (int i = 0; i < maxn; i++){l[i] = INF;r[i] = -INF;t[i] = 0;}for (int i = 0; i < n; i++){int p; scanf("%d", &p);l[p] = min(l[p], i);r[p] = max(r[p], i);t[p]++;}for (int i = 0; i < maxn; i++)ma = max(ma, t[i]);for (int i = 0; i < maxn; i++)if (ma == t[i] && ans > r[i] - l[i] + 1){ans = r[i] - l[i] + 1; j = i;}printf("%d %d\n", l[j] + 1, r[j] + 1);}return 0;}