17-比赛1 F

Seg-El has last chance to make the final changes in order to prevent the destruction of Krypton.

He is provided an array Arr[ ] of N elements.

For each element Arr [ i ], he needs to find the largest element Arr [ j ] where j < i and Arr [ j ] < Arr [ i ]

Help him to achieve this task.

Input

First line comprises N- the number of elements in the array.

Second line contains N space separated integers denoting the elements of the array.

Output

Print N lines denoting the required answer as specified above. In case no such value exists for some

Arr [ i ], print "-1" (without quotes).

Constraints

• 1 ≤ N ≤ 200000
• 1 ≤ Arr [ i ] ≤ 10¹⁵

Example

5 
1 2 3 5 4 

Output:

-1 
1 
2 
3 
3 

通过 set 或 map 就可轻松解决；

关键字 ：  set ，map ,  lower_bound()函数

1.学长的代码

# include <bits/stdc++.h>
using namespace std;
set<long long> s;
int main ()
{
    int n;
    long long x;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &x);
        x = -x;
        auto iter = s.lower_bound(x + 1);
        if (iter == s.end()) puts("-1");
        else printf("%lld
", -*iter);
        s.insert(x);
    }
    return 0;
}

2 . 一开始不会使用set ，用map自己做的

#include<bits/stdc++.h>
using namespace std;
int main()
{
    map<long long,int> imap;
    long long t;
    int n;
    cin>>n;
    for(int i =1;i<=n;i++)
    {
        scanf("%lld",&t);
        t = -t;
        //map<long long,int>::iterator it;
        auto it = imap.lower_bound(t+1);
        if(it == imap.end()) cout<<"-1"<<endl;
        else
           cout<<-(it->first)<<endl;
        imap[t] = i;
    }
    return 0;
}