计算给定素数 P 等于自身时具有模逆的数组元素
原文:https://www . geesforgeks . org/count-array-elements-having-modular-inverse-欠给定质数-p-equal-to-self/
给定一个大小为 N 的数组arr【】和一个质数 P ，任务是对数组元素进行计数，使得在模 P 下元素的模乘逆等于元素本身。
示例:
输入: arr[] = {1，6，4，5}，P = 7 输出: 2 解释: 模 P(= 7)下 arr0的模乘逆本身就是 arr0。 模 P(= 7)下 arr1的模乘逆本身就是 arr1。 因此，要求输出为 2。
输入: arr[] = {1，3，8，12，12}，P = 13 T3】输出: 3
天真法:解决这个问题最简单的方法就是遍历数组并打印数组元素的计数，使得模 P 下元素的模乘逆等于元素本身。
时间复杂度: O(N * log P) 辅助空间: O(1)
高效方法:为了优化上述方法，该想法基于以下观察:
如果 X 和 Y 是两个数字使得 (X × Y) % P = 1 ，那么 Y 是 X 的模逆。 所以如果 Y 是 X 本身，那么 (X × X) % P 一定是 1。
按照以下步骤解决问题:
- 初始化一个变量，比如 cntElem 来存储满足给定条件的元素计数。
- 遍历给定数组，检查 (arr[i] * arr[i]) % P 是否等于 1 。如果发现为真，则将计数增加1。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to get the count
// of elements that satisfy
// the given condition.
int equvInverse(int arr[],
int N, int P)
{
// Stores count of elements
// that satisfy the condition
int cntElem = 0;
// Traverse the given array.
for (int i = 0; i < N; i++) {
// If square of current
// element is equal to 1
if ((arr[i] * arr[i]) % P
== 1) {
cntElem++;
}
}
return cntElem;
}
// Driver Code
int main()
{
int arr[] = { 1, 6, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
int P = 7;
cout << equvInverse(arr, N, P);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr,
int N, int P)
{
// Stores count of elements
// that satisfy the condition
int cntElem = 0;
// Traverse the given array.
for(int i = 0; i < N; i++)
{
// If square of current
// element is equal to 1
if ((arr[i] * arr[i]) % P == 1)
{
cntElem++;
}
}
return cntElem;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 6, 4, 5 };
int N = arr.length;
int P = 7;
System.out.println(equvInverse(arr, N, P));
}
}
// This code is contributed by akhilsaini
Python 3
# Python3 program to implement
# the above approach
# Function to get the count
# of elements that satisfy
# the given condition.
def equvInverse(arr, N, P):
# Stores count of elements
# that satisfy the condition
cntElem = 0
# Traverse the given array.
for i in range(0, N):
# If square of current
# element is equal to 1
if ((arr[i] * arr[i]) % P == 1):
cntElem = cntElem + 1
return cntElem
# Driver Code
if __name__ == "__main__":
arr = [ 1, 6, 4, 5 ]
N = len(arr)
P = 7
print(equvInverse(arr, N, P))
# This code is contributed by akhilsaini
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr, int N, int P)
{
// Stores count of elements
// that satisfy the condition
int cntElem = 0;
// Traverse the given array.
for(int i = 0; i < N; i++)
{
// If square of current
// element is equal to 1
if ((arr[i] * arr[i]) % P == 1)
{
cntElem++;
}
}
return cntElem;
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 6, 4, 5 };
int N = arr.Length;
int P = 7;
Console.WriteLine(equvInverse(arr, N, P));
}
}
// This code is contributed by akhilsaini
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to get the count
// of elements that satisfy
// the given condition.
function equvInverse(arr, N, P)
{
// Stores count of elements
// that satisfy the condition
let cntElem = 0;
// Traverse the given array.
for (let i = 0; i < N; i++) {
// If square of current
// element is equal to 1
if ((arr[i] * arr[i]) % P
== 1) {
cntElem++;
}
}
return cntElem;
}
// Driver Code
let arr = [ 1, 6, 4, 5 ];
let N = arr.length;
let P = 7;
document.write(equvInverse(arr, N, P));
// This code is contributed by subham348.
</script>
Output:
2
时间复杂度:O(N) T5辅助空间:** O(1)
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