Codeforces Round #450 (Div. 2) C. Remove Extra One【*模拟链表/一个数比前面所有数大就是个record。删掉一个数,让record的个数尽量多。】

时间:2023-03-09 17:20:44
Codeforces Round #450 (Div. 2) C. Remove Extra One【*模拟链表/一个数比前面所有数大就是个record。删掉一个数,让record的个数尽量多。】
C. Remove Extra One
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.

We remind that in a sequence of numbers a1, a2, ..., ak the element ai is a record if for every integer j (1 ≤ j < i) the following holds:aj < ai.

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the length of the permutation.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation. All the integers are distinct.

Output

Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples
input
1
1
output
1
input
5
5 1 2 3 4
output
5
Note

In the first example the only element can be removed.

【代码】:

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+;
int mx,mx2,n,mn=-1e9,res;
int a[N];
int main()
{
scanf("%d",&n);
for(int i=;i<n;i++)
{
int x;
scanf("%d",&x);
if(x>mx)
{
mx2=mx;
mx=max(mx,x);
a[x]--;
}
else if(x>mx2)
{
a[mx]++;
mx2=max(mx2,x);
}
}
for(int i=;i<=n;i++)
{
if(a[i]>mn)
{
mn=a[i];
res=i;
}
}
cout<<res<<endl;
return ;
}