Codeforces 558B Amr and The Large Array

最新推荐文章于 2019-05-29 12:24:53 发布

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最新推荐文章于 2019-05-29 12:24:53 发布

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文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/u014028317/article/details/46898439

本文介绍了一个算法问题，即如何从一个大型数组中找到一个子数组，使得该子数组的元素出现频率最高次数与原数组相同，并且子数组尽可能短小。文章通过示例解释了问题并提供了具体的输入输出样例。

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 a_i (1 ≤ a_i ≤ 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 B_i = A_{l + i - 1} for all 1 ≤ i ≤ r - l + 1

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;

struct node
{
    int num;
    int time;
    int l;
    int r;
    int range;
} index[1000010];

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n;
    while(cin>>n)
    {
        map<int, int>g;
        g.clear();
        memset(index, 0, sizeof(index));
        int t;
        int d = 0;
        int dd;
        int maxd = 1;
        int maxtime = 0;
        int maxrange = 1;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &t);
            if(g[t] != 0)
            {
                dd = g[t];
                index[dd].time++;
                index[dd].r = i;
                index[dd].range = index[dd].r-index[dd].l+1;
                if(maxtime<index[dd].time || maxtime==index[dd].time&&maxrange>index[dd].range)
                {
                    maxtime = index[dd].time;
                    maxd = dd;
                    maxrange = index[dd].range;
                }
            }
            else
            {
                g[t] = ++d;
                dd = g[t];
                index[dd].num = t;
                index[dd].time = 1;
                index[dd].l = index[dd].r = i;
                index[dd].range = 1;
                if(maxtime < index[dd].time)
                {
                    maxtime = index[dd].time;
                    maxd = dd;
                    maxrange = 1;
                }
            }
        }
        printf("%d %d\n", index[maxd].l, index[maxd].r);
    }
    return 0;
}