按 X 除时数组元素商的非递减顺序打印索引
原文:https://www . geeksforgeeks . org/print-indexes-in-非递减顺序-按 x 除的数组元素商/
给定一个由 N 个整数和一个整数 X 组成的数组arr【】】，任务是通过 X 对数组元素执行整数除法，并按照获得的商的非递减顺序打印数组的索引。
示例:
输入: N = 3，X = 3，顺序[] = {2，7，4} 输出: 1 3 2 解释:** 将数组元素除以 3 后，数组修改为{0，2，1}。因此，所需的输出顺序是 1 3 2。
输入: N = 5，X = 6，顺序[] = {9，10，4，7，2} 输出: 3 5 1 2 4 解释:** 将数组元素除以 6 后，将数组元素修改为 1 1 0 1 0。因此，所需的序列是 3 5 1 2 4。
方法:按照以下步骤解决问题:
- 遍历数组
- 初始化对的向量。
- 对于每个数组元素，将除以 X 得到的商的值存储为向量中的对的第一个元素，将第二个元素存储为所需顺序中整数的位置。
- 遍历结束后，对向量进行排序，最后打印出对的所有第二个元素。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the order of array
// elements generating non-decreasing
// quotient after division by X
void printOrder(int order[], int N, int X)
{
// Stores the quotient and the order
vector<pair<int, int> > vect;
// Traverse the array
for (int i = 0; i < N; i++) {
if (order[i] % X == 0) {
vect.push_back({ order[i] / X,
i + 1 });
}
else {
vect.push_back({ order[i] / X + 1,
i + 1 });
}
}
// Sort the vector
sort(vect.begin(), vect.end());
// Print the order
for (int i = 0; i < N; i++) {
cout << vect[i].second << " ";
}
cout << endl;
}
// Driver Code
int main()
{
int N = 3, X = 3;
int order[] = { 2, 7, 4 };
printOrder(order, N, X);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
// Function to print the order of array
// elements generating non-decreasing
// quotient after division by X
static void printOrder(int order[], int N, int X)
{
// Stores the quotient and the order
ArrayList<int[]> vect = new ArrayList<>();
// Traverse the array
for (int i = 0; i < N; i++) {
if (order[i] % X == 0) {
vect.add(new int[] { order[i] / X, i + 1 });
}
else {
vect.add(
new int[] { order[i] / X + 1, i + 1 });
}
}
// Sort the vector
Collections.sort(vect, (a, b) -> a[0] - b[0]);
// Print the order
for (int i = 0; i < N; i++) {
System.out.print(vect.get(i)[1] + " ");
}
System.out.println();
}
// Driver Code
public static void main(String args[])
{
int N = 3, X = 3;
int order[] = { 2, 7, 4 };
printOrder(order, N, X);
}
}
// This code is contributed by hemanth gadarla
Python 3
# Python3 program for the above approach
# Function to print the order of array
# elements generating non-decreasing
# quotient after division by X
def printOrder(order, N, X):
# Stores the quotient and the order
vect = []
# Traverse the array
for i in range(N):
if (order[i] % X == 0):
vect.append([order[i] // X, i + 1])
else:
vect.append([order[i] // X + 1,i + 1])
# Sort the vector
vect = sorted(vect)
# Print the order
for i in range(N):
print(vect[i][1], end = " ")
# Driver Code
if __name__ == '__main__':
N, X = 3, 3
order = [2, 7, 4]
printOrder(order, N, X)
# This code is contributed by mohit kumar 29
C
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to print the order of array
// elements generating non-decreasing
// quotient after division by X
static void printOrder(int[] order, int N, int X)
{
// Stores the quotient and the order
List<Tuple<int,
int>> vect = new List<Tuple<int,
int>>();
// Traverse the array
for (int i = 0; i < N; i++)
{
if (order[i] % X == 0)
{
vect.Add(new Tuple<int,int>((order[i] / X), i + 1));
}
else
{
vect.Add(new Tuple<int,int>((order[i] / X + 1), i + 1));
}
}
// Sort the vector
vect.Sort();
// Print the order
for (int i = 0; i < N; i++)
{
Console.Write(vect[i].Item2 + " ");
}
Console.WriteLine();
}
// Driver Code
public static void Main()
{
int N = 3, X = 3;
int[] order = { 2, 7, 4 };
printOrder(order, N, X);
}
}
// This code is contributed by code_hunt.
java 描述语言
<script>
// Javascript program for the above approach
// Function to print the order of array
// elements generating non-decreasing
// quotient after division by X
function printOrder(order, N, X)
{
// Stores the quotient and the order
let vect = [];
// Traverse the array
for(let i = 0; i < N; i++)
{
if (order[i] % X == 0)
{
vect.push([ order[i] / X, i + 1 ]);
}
else
{
vect.push([ order[i] / X + 1, i + 1 ]);
}
}
// Sort the vector
vect.sort(function(a, b){return a[0] - b[0]});
// Print the order
for(let i = 0; i < N; i++)
{
document.write(vect[i][1] + " ");
}
document.write();
}
// Driver Code
let N = 3, X = 3;
let order = [ 2, 7, 4 ];
printOrder(order, N, X);
// This code is contributed by unknown2108
</script>
Output:
1 3 2
时间复杂度: O(N) 辅助空间: O(N)
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