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Zero-complexity Transposition

[编程题]Zero-complexity Transposition

热度指数：12805 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

You are given a sequence of integer numbers. Zero-complexity transposition of the sequence is the reverse of this sequence. Your task is to write a program that prints zero-complexity transposition of the given sequence.

输入描述:

For each case, the first line of the input file contains one integer n-length of the sequence (0 ＜ n ≤ 10 000). The second line contains n integers numbers-a1, a2, …, an (-1 000 000 000 000 000 ≤ ai ≤ 1 000 000 000 000 000).

输出描述:

For each case, on the first line of the output file print the sequence in the reverse order.

示例1

输入

5
-3 4 6 -8 9

输出

9 -8 6 4 -3

算法知识视频讲解

Java

WIT_221

import java.util.Scanner;

//双指针的思路

public class Main {

public static void main(String[] args) {

Scanner in = new Scanner(System.in);

// 注意 hasNext 和 hasNextLine 的区别

int len = in.nextInt();

int[] nums = new int[len];

for(int i = 0 ;i<len;i++){

nums[i] = in.nextInt();

}

int left = 0 ;

int right = len-1;

while(left<=right){

int temp = nums[left];

nums[left] = nums[right];

nums[right] = temp;

left++;

right--;

}

for(int i:nums){

System.out.print(i+" ");

}

发表于 2024-03-17 10:48:32 回复(0)

Java

早豆浆晚牛奶

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s;
        while ((s = br.readLine()) != null) {
            int n = Integer.parseInt(s);
            String[] ss = br.readLine().split(" ");

            for (int i = n - 1; i >= 0; --i) {
                System.out.print(ss[i] + " ");
            }

        }
    }
}

发表于 2021-04-02 14:46:49 回复(0)