17-比赛1 F - 较小元素

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).

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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；

1.学长的代码

 1 /*
   从前往后遍历数组，维护一个setset，对于数组的当前元素acurrentacurrent，使用setset的lower_boundlower_bound函数查找
   setset里面满足≥acurrent≥acurrent的元素里的最小元素，再把acurrentacurrent插入setset里面就行了。
 2 但是题目问的是当前元素前面的元素里满足<acurrent<acurrent的元素里的最大元素。
 3 怎么办？
 4 首先把数组里的每个元素取个相反数就行了。
   */
 5 # include <bits/stdc++.h>
 6 using namespace std;
 7 set<long long> s;
 8 int main ()
 9 {
10     int n;
11     long long x;
12     scanf("%d", &n);
13     for (int i = 1; i <= n; ++i) {
14         scanf("%lld", &x);
15         x = -x;
16         auto iter = s.lower_bound(x + 1);
17         if (iter == s.end()) puts("-1");
18         else printf("%lld\n", -*iter);
19         s.insert(x);
20     }
21     return 0;
22 }

2 . 一开始不会使用set ，用map自己做的

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     map<long long,int> imap;
 6     long long t;
 7     int n;
 8     cin>>n;
 9     for(int i =1;i<=n;i++)
10     {
11         scanf("%lld",&t);
12         t = -t;
13         //map<long long,int>::iterator it;
14         auto it = imap.lower_bound(t+1);
15         if(it == imap.end()) cout<<"-1"<<endl;
16         else
17            cout<<-(it->first)<<endl;
18         imap[t] = i;
19     }
20     return 0;
21 }