CF 558B Amr and The Large Array（暴力法）

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

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

有一个序列，找出包含它的出现次数最多的那个数的所有的那些次中最小的序列。就是找出它的众数，子序列的众数出现的次数和原序列的次数一样，

求最短的这个子序列。

当时眼瞎，没看到最短这个条件。

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>using namespace std;#define maxn 100005#define maxa 1000005struct count_a{int value, min, max;count_a(){value = 0; min = 0; max = 0;}};count_a cnt[maxa];int num[maxn];int main(){//取最短区间int n;int max_num = 0, last_value = 0, last_dis = maxn;scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d", &num[i]);cnt[num[i]].value++;if (cnt[num[i]].min != 0){ cnt[num[i]].max = i; }else{cnt[num[i]].min = i;cnt[num[i]].max = i;}if (cnt[num[i]].value > last_value){last_value = cnt[num[i]].value;max_num = num[i];last_dis = cnt[num[i]].max - cnt[num[i]].min;}else if (cnt[num[i]].value == last_value){if (cnt[num[i]].max - cnt[num[i]].min < last_dis){last_value = cnt[num[i]].value;max_num = num[i];last_dis = cnt[num[i]].max - cnt[num[i]].min;}}}printf("%d %d", cnt[max_num].min, cnt[max_num].max);//system("pause");return 0;}