给定一个大小为N的数组arr[] ，任务是检查是否可以通过多次旋转数组元素的数字来将所有数组元素转换为 pronic 数。
例子：
Input: {321, 402, 246, 299}
Output: True
Explanation:
arr[0] → Right rotation once modifies arr[0] to 132 (= 11 × 12).
arr[1] → Right rotation once modifies arr[0] to 240 (= 15 × 16).
arr[2] → Right rotation twice modifies arr[2] to 462 (= 21 × 22).
arr[3] → Right rotation twice modifies arr[3] to 992 (= 31 × 32).
Input: {433, 653, 402, 186}
Output: False
处理方法：按照以下步骤解决问题：
- 遍历数组并检查每个数组元素，是否可以将其转换为 pronic 数。
- 对于每个数组元素，应用所有可能的旋转并在每次旋转后检查生成的数字是否是 pronic 的。
- 如果无法将任何数组元素转换为 pronic 数，则打印“False” 。
- 否则，打印“True” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// function to check Pronic Number
bool isPronic(int x)
{
for (int i = 0; i < (int)(sqrt(x)) + 1; i++)
{
// Checking Pronic Number
// by multiplying consecutive
// numbers
if (x == i * (i + 1))
{
return true;
}
}
return false;
}
// Function to check if any permutation
// of val is a pronic number or not
bool checkRot(int val)
{
string temp = to_string(val);
for (int i = 0; i < temp.length(); i++)
{
if (isPronic(stoi(temp)) == true)
{
return true;
}
temp = temp.substr(1, temp.size() - 1) + temp[0];
}
return false;
}
// Function to check if all array
// elements can be converted to
// a pronic number or not
bool check(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
{
// If current element
// cannot be converted
// to a pronic number
if (checkRot(arr[i]) == false)
{
return false;
}
}
return true;
}
// Driven Program
int main()
{
// Given array
int arr[] = { 321, 402, 246, 299 };
int N = sizeof(arr) / sizeof(arr[0]);
// function call
cout << (check(arr, N) ? "True" : "False");
return 0;
}
// This code is contributed by Kingash. Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// function to check Pronic Number
static boolean isPronic(int x)
{
for (int i = 0; i < (int)(Math.sqrt(x)) + 1; i++) {
// Checking Pronic Number
// by multiplying consecutive
// numbers
if (x == i * (i + 1)) {
return true;
}
}
return false;
}
// Function to check if any permutation
// of val is a pronic number or not
static boolean checkRot(int val)
{
String temp = Integer.toString(val);
for (int i = 0; i < temp.length(); i++)
{
if (isPronic(Integer.parseInt(temp)) == true) {
return true;
}
temp = temp.substring(1) + temp.charAt(0);
}
return false;
}
// Function to check if all array
// elements can be converted to
// a pronic number or not
static boolean check(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
{
// If current element
// cannot be converted
// to a pronic number
if (checkRot(arr[i]) == false)
{
return false;
}
}
return true;
}
// Driver code
public static void main(String[] args)
{
// Given array
int arr[] = { 321, 402, 246, 299 };
int N = arr.length;
// Function call
System.out.println(
(check(arr, N) ? "True" : "False"));
}
}
// This code is contributed by Kingash.Python3
# Python implementation of
# the above approach
# Function to check if a number
# is a pronic number or not
def isPronic(n):
for i in range(int(n**(1 / 2)) + 1):
if i * (i + 1) == n:
return True
return False
# Function to check if any permutation
# of n is a pronic number or not
def checkRot(n):
temp = str(n)
for i in range(len(temp)):
if isPronic(int(temp)):
return True
temp = temp[1:]+temp[0]
return False
# Function to check if all array
# elements can be converted to
# a pronic number or not
def check(arr):
# Traverse the array
for i in arr:
# If current element
# cannot be converted
# to a pronic number
if not checkRot(i):
return False
return True
# Driver Code
arr = [ 321, 402, 246, 299 ]
print(check(arr))C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// function to check Pronic Number
static bool isPronic(int x)
{
int val = (int)Math.Sqrt(x);
val += 1;
for (int i = 0; i < val; i++)
{
// Checking Pronic Number
// by multiplying consecutive
// numbers
if (x == i * (i + 1))
{
return true;
}
}
return false;
}
// Function to check if any permutation
// of val is a pronic number or not
static bool checkRot(int val)
{
string temp = val.ToString();
for (int i = 0; i < temp.Length; i++)
{
int a = Int32.Parse(temp);
if (isPronic(a) == true)
{
return true;
}
temp = temp.Substring(1, temp.Length - 1) + temp[0];
}
return false;
}
// Function to check if all array
// elements can be converted to
// a pronic number or not
static bool check(int []arr, int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
{
// If current element
// cannot be converted
// to a pronic number
if (checkRot(arr[i]) == false)
{
return false;
}
}
return true;
}
// Driven Program
public static void Main()
{
// Given array
int []arr = { 321, 402, 246, 299 };
int N = arr.Length;
// function call
Console.WriteLine(check(arr, N) ? "True" : "False");
}
}
// This code is contributed by ipg2016107.Javascript
输出：
True时间复杂度： O(N 3/2 )
辅助空间： O(1)
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