B. Amr and The Large Array(Codeforces Round #312 (Div. 2)+找出现次数最多且区间最小)

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

   题意：给出一个序列，然后找出出现次数最多，但区间占用长度最短的区间左右值。

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
 
#define N 1000001
 
using namespace std;
 
int n,m;
 
struct node
{
    int x;
    int y;
    int ans;
    int cnt;
}q[1000100];
 
bool cmp(node a,node b)
{
    if(a.cnt == b.cnt)
    {
        return a.ans < b.ans;
    }
    return a.cnt > b.cnt;
}
 
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int mm;
        for(int i=0;i<=N;i++)
        {
            q[i].cnt = 0;
        }
        int maxx = 0;
        int a;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a);
            if(mm<a)
            {
                mm = a;
            }
            if(q[a].cnt == 0)
            {
                q[a].x = i;
                q[a].y = i;
                q[a].ans = 0;
                q[a].cnt++;
            }
            else
            {
                q[a].cnt++;
                q[a].y = i;
                q[a].ans = q[a].y - q[a].x;
            }
            if(maxx<q[a].cnt)
            {
                maxx = q[a].cnt;
            }
        }
        sort(q,q+N,cmp);
        printf("%d %d\n",q[0].x,q[0].y);
 
    }
    return 0;
}