CF 558B(Amr and The Large Array-计数)

最新推荐文章于 2019-05-29 12:24:53 发布

原创最新推荐文章于 2019-05-29 12:24:53 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

杂题专栏收录该内容

50 篇文章

订阅专栏

本博客探讨了一个算法问题：如何在给定数组中找到一个最小子段，使得其最大重复元素次数与原始数组相同。通过输入数组大小和元素，输出能够保持数组美丽性的最小子段边界。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

对每个可能值计数

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (2000000+10)
#define N (1000000)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int l[MAXN],r[MAXN],t[MAXN],n;
int main()
{
//	freopen("B.in","r",stdin);
//	freopen(".out","w",stdout);
	
	For(i,N) l[i]=INF,r[i]=-INF,t[i]=0;
	
	cin>>n;
	For(i,n)
	{
		int p;scanf("%d",&p);
		l[p]=min(l[p],i);
		r[p]=max(r[p],i);
		t[p]++;
	}
	
	int ma=0,ans=INF,j=0;
	For(i,N) ma=max(ma,t[i]);
	For(i,N)
		if (ma==t[i]&&ans>r[i]-l[i]+1)
		{
			 ans=r[i]-l[i]+1;j=i;
		}
	cout<<l[j]<<' '<<r[j]<<endl;
	
	return 0;
}