给定一个由N个整数和一个整数X组成的数组arr [] ,任务是用X对数组元素进行整数除法,并以得到的商的非降序打印该数组的索引。
例子:
Input: N = 3, X = 3, order[] = {2, 7, 4}
Output: 1 3 2
Explanation:
After dividing the array elements by 3, the array modifies to {0, 2, 1}. Therefore, required order of output is 1 3 2.
Input: N = 5, X = 6, order[] = {9, 10, 4, 7, 2}
Output: 3 5 1 2 4
Explanation:
After dividing the array elements by 6, the array elements modify to 1 1 0 1 0. Therefore, the required sequence is 3 5 1 2 4.
方法:请按照以下步骤解决问题:
- 遍历数组
- 初始化向量对。
- 对于每个数组元素,将按X除后获得的商的值存储为向量中对的第一个元素,将第二个元素存储为整数(按要求的顺序)。
- 遍历之后,对向量进行排序,然后最终打印对中的所有第二个元素。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
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 > 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
/*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 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
Python3
# 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> vect = new List>();
// Traverse the array
for (int i = 0; i < N; i++)
{
if (order[i] % X == 0)
{
vect.Add(new Tuple((order[i] / X), i + 1));
}
else
{
vect.Add(new Tuple((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.
输出:
1 3 2
时间复杂度: O(N)
辅助空间: O(N)