小易有一个长度为n的整数序列,a_1,...,a_n。然后考虑在一个空序列b上进行n次以下操作:
1、将a_i放入b序列的末尾
2、逆置b序列
小易需要你计算输出操作n次之后的b序列。
输入包括两行,第一行包括一个整数n(2 ≤ n ≤ 2*10^5),即序列的长度。 第二行包括n个整数a_i(1 ≤ a_i ≤ 10^9),即序列a中的每个整数,以空格分割。
在一行中输出操作n次之后的b序列,以空格分割,行末无空格。
4 1 2 3 4
4 2 1 3
//不知道楼上整链表干啥 #include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; vector<int> a; for(int i = 0; i<n; i++) { int temp; cin>>temp; a.push_back(temp); }; int i; for(i=n-1;i>=0;i-=2) { cout<<a[i]<<' '; } if(n%2 == 0) for(int i = 0; i<n; i+=2) cout<<a[i]<<' '; else for(int i = 1; i<n;i+=2) cout<<a[i]<<' '; //cin>>n; return 0; }
import java.util.Scanner; import java.util.LinkedList; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); LinkedList<Integer> list = new LinkedList<>(); boolean left = false; for (int i = 0; i < n; i++) { if (left) { list.addFirst(scanner.nextInt()); } else { list.addLast(scanner.nextInt()); } left = !left; } if (!left) { for (Integer i : list) { System.out.print(i + " "); } } else { while (!list.isEmpty()) { System.out.print(list.removeLast() + " "); } } } }
#include<iostream> #include<deque> using namespace std; int main(){ int n; cin>>n; deque<int> a; for(int i=0;i<n;i++){ int x; cin>>x; if(i%2==0) a.push_back(x); else a.push_front(x); } if(n%2==0){ while(!a.empty()){ cout<<a.front()<<" "; a.pop_front(); } cout<<endl; }else{ while(!a.empty()){ cout<<a.back()<<" "; a.pop_back(); } cout<<endl; } return 0; }
#include <iostream> #include <vector> #include <string> using namespace std; void reverseArr(vector<int>& arr) { int size = arr.size(); vector<int> res(size, 0); int head = 0; int tail = size - 1; bool isHead = true; for (int i = tail; i >= 0; i--) { int val = arr[i]; if (isHead) { res[head] = val; head++; } else { res[tail] = val; tail--; } isHead ^= 1; } for (int i = 0; i < size; i++) { cout << res[i]; if (i != size - 1) { cout << " "; } } cout << endl; return; } vector<string> split(string str, string pattern) { string::size_type pos; vector<string> result; str += pattern; int size = str.size(); for (int i = 0; i<size; i++) { pos = str.find(pattern, i); if (pos<size) { std::string s = str.substr(i, pos - i); result.push_back(s); i = pos + pattern.size() - 1; } } return result; } int main() { string s; getline(cin, s); getline(cin, s); vector<string> tmp = split(s, " "); vector<int> input; for (int i = 0; i < tmp.size(); i++) { input.push_back(atoi(tmp[i].c_str())); } reverseArr(input); return 0; }