求模 p |集合 1 下的平方根(当 p 是 4*i + 3 的形式时)
原文:https://www . geeksforgeeks . org/find-求模下平方根-p-set-1-当-p 是-4i-3 的形式时/
给定一个数“n”和一个素数“p”，求模 p 下 n 的平方根，如果它存在的话。可以给出 p 是 4i + 3 (OR p % 4 = 3)的形式，其中 I 是整数。这类素数的例子有 7、11、19、23、31、…等 例子:*
Input: n = 2, p = 7
Output: 3 or 4
3 and 4 both are square roots of 2 under modulo
7 because (3*3) % 7 = 2 and (4*4) % 7 = 2
Input: n = 2, p = 5
Output: Square root doesn't exist
天真解:尝试从 2 到 p-1 的所有数字。对于每个数字 x，检查 x 是否是模 p 下 n 的平方根
C++
// A Simple C++ program to find square root under modulo p
// when p is 7, 11, 19, 23, 31, ... etc,
#include <iostream>
using namespace std;
// Returns true if square root of n under modulo p exists
void squareRoot(int n, int p)
{
n = n % p;
// One by one check all numbers from 2 to p-1
for (int x = 2; x < p; x++) {
if ((x * x) % p == n) {
cout << "Square root is " << x;
return;
}
}
cout << "Square root doesn't exist";
}
// Driver program to test
int main()
{
int p = 7;
int n = 2;
squareRoot(n, p);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Simple Java program to find square
// root under modulo p when p is 7,
// 11, 19, 23, 31, ... etc,
import java .io.*;
class GFG {
// Returns true if square root of n
// under modulo p exists
static void squareRoot(int n, int p)
{
n = n % p;
// One by one check all numbers
// from 2 to p-1
for (int x = 2; x < p; x++) {
if ((x * x) % p == n) {
System.out.println("Square "
+ "root is " + x);
return;
}
}
System.out.println("Square root "
+ "doesn't exist");
}
// Driver Code
public static void main(String[] args)
{
int p = 7;
int n = 2;
squareRoot(n, p);
}
}
// This code is contributed by Anuj_67
Python 3
# A Simple Python program to find square
# root under modulo p when p is 7, 11,
# 19, 23, 31, ... etc,
# Returns true if square root of n under
# modulo p exists
def squareRoot(n, p):
n = n % p
# One by one check all numbers from
# 2 to p-1
for x in range (2, p):
if ((x * x) % p == n) :
print( "Square root is ", x)
return
print( "Square root doesn't exist")
# Driver program to test
p = 7
n = 2
squareRoot(n, p)
# This code is Contributed by Anuj_67
C
// A Simple C# program to find square
// root under modulo p when p is 7,
// 11, 19, 23, 31, ... etc,
using System;
class GFG {
// Returns true if square root of n
// under modulo p exists
static void squareRoot(int n, int p)
{
n = n % p;
// One by one check all numbers
// from 2 to p-1
for (int x = 2; x < p; x++) {
if ((x * x) % p == n) {
Console.Write("Square "
+ "root is " + x);
return;
}
}
Console.Write("Square root "
+ "doesn't exist");
}
// Driver Code
static void Main()
{
int p = 7;
int n = 2;
squareRoot(n, p);
}
}
// This code is contributed by Anuj_67
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A Simple PHP program to find
// square root under modulo p
// when p is 7, 11, 19, 23, 31,
// ... etc,
// Returns true if square
// root of n under modulo
// p exists
function squareRoot($n, $p)
{
$n = $n % $p;
// One by one check all
// numbers from 2 to p-1
for ($x = 2; $x < $p; $x++)
{
if (($x * $x) % $p == $n)
{
echo("Square root is " . $x);
return;
}
}
echo("Square root doesn't exist");
}
// Driver Code
$p = 7;
$n = 2;
squareRoot($n, $p);
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// A Simple Javascript program to find square
// root under modulo p when p is 7,
// 11, 19, 23, 31, ... etc,
// Returns true if square root of n
// under modulo p exists
function squareRoot(n,p)
{
n = n % p;
// One by one check all numbers
// from 2 to p-1
for (let x = 2; x < p; x++) {
if ((x * x) % p == n) {
document.write("Square "
+ "root is " + x);
return;
}
}
document.write("Square root "
+ "doesn't exist");
}
// Driver Code
let p = 7;
let n = 2;
squareRoot(n, p);
// This code is contributed by rag2127
</script>
输出:
Square root is 3
这个解的时间复杂度是 O(p) 直接法:如果 p 是 4*i + 3 的形式，那么就存在求平方根的快速方法。
If n is in the form 4*i + 3 with i >= 1 (OR p % 4 = 3)
And
If Square root of n exists, then it must be
±n<sup>(p + 1)/4</sup>
以下是上述思路的实现:
C++
// An efficient C++ program to find square root under
// modulo p when p is 7, 11, 19, 23, 31, ... etc.
#include <iostream>
using namespace std;
// Utility function to do modular exponentiation.
// It returns (x^y) % p.
int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns true if square root of n under modulo p exists
// Assumption: p is of the form 3*i + 4 where i >= 1
void squareRoot(int n, int p)
{
if (p % 4 != 3) {
cout << "Invalid Input";
return;
}
// Try "+(n^((p + 1)/4))"
n = n % p;
int x = power(n, (p + 1) / 4, p);
if ((x * x) % p == n) {
cout << "Square root is " << x;
return;
}
// Try "-(n ^ ((p + 1)/4))"
x = p - x;
if ((x * x) % p == n) {
cout << "Square root is " << x;
return;
}
// If none of the above two work, then
// square root doesn't exist
cout << "Square root doesn't exist ";
}
// Driver program to test
int main()
{
int p = 7;
int n = 2;
squareRoot(n, p);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// An efficient Java program to find square root under
// modulo p when p is 7, 11, 19, 23, 31, ... etc.
public class GFG {
// Utility function to do modular exponentiation.
// It returns (x^y) % p.
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y %2== 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns true if square root of n under modulo p exists
// Assumption: p is of the form 3*i + 4 where i >= 1
static void squareRoot(int n, int p)
{
if (p % 4 != 3) {
System.out.print("Invalid Input");
return;
}
// Try "+(n^((p + 1)/4))"
n = n % p;
int x = power(n, (p + 1) / 4, p);
if ((x * x) % p == n) {
System.out.print("Square root is " + x);
return;
}
// Try "-(n ^ ((p + 1)/4))"
x = p - x;
if ((x * x) % p == n) {
System.out.print("Square root is " + x);
return;
}
// If none of the above two work, then
// square root doesn't exist
System.out.print("Square root doesn't exist ");
}
// Driver program to test
static public void main(String[] args) {
int p = 7;
int n = 2;
squareRoot(n, p);
}
}
Python 3
# An efficient python3 program to find square root
# under modulo p when p is 7, 11, 19, 23, 31, ... etc.
# Utility function to do modular exponentiation.
# It returns (x^y) % p.
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more
# than or equal to p
while (y > 0):
# If y is odd, multiply x with result
if (y & 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# Returns true if square root of n under
# modulo p exists. Assumption: p is of the
# form 3*i + 4 where i >= 1
def squareRoot(n, p):
if (p % 4 != 3) :
print( "Invalid Input" )
return
# Try "+(n^((p + 1)/4))"
n = n % p
x = power(n, (p + 1) // 4, p)
if ((x * x) % p == n):
print( "Square root is ", x)
return
# Try "-(n ^ ((p + 1)/4))"
x = p - x
if ((x * x) % p == n):
print( "Square root is ", x )
return
# If none of the above two work, then
# square root doesn't exist
print( "Square root doesn't exist " )
# Driver Code
p = 7
n = 2
squareRoot(n, p)
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
C
// An efficient C# program to find square root under
// modulo p when p is 7, 11, 19, 23, 31, ... etc.
using System;
public class GFG {
// Utility function to do modular exponentiation.
// It returns (x^y) % p.
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y %2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns true if square root of n under modulo p exists
// Assumption: p is of the form 3*i + 4 where i >= 1
static void squareRoot(int n, int p)
{
if (p % 4 != 3) {
Console.Write("Invalid Input");
return;
}
// Try "+(n^((p + 1)/4))"
n = n % p;
int x = power(n, (p + 1) / 4, p);
if ((x * x) % p == n) {
Console.Write("Square root is " + x);
return;
}
// Try "-(n ^ ((p + 1)/4))"
x = p - x;
if ((x * x) % p == n) {
Console.Write("Square root is " + x);
return;
}
// If none of the above two work, then
// square root doesn't exist
Console.Write("Square root doesn't exist ");
}
// Driver program to test
static public void Main() {
int p = 7;
int n = 2;
squareRoot(n, p);
}
}
// This code is contributed by Ita_c.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// An efficient PHP program
// to find square root under
// modulo p when p is 7, 11,
// 19, 23, 31, ... etc.
// Utility function to do
// modular exponentiation.
// It returns (x^y) % p.
function power($x, $y, $p)
{
// Initialize result
$res = 1;
// Update x if it
// is more than or
// equal to p
$x = $x % $p;
while ($y > 0)
{
// If y is odd, multiply
// x with result
if ($y & 1)
$res = ($res * $x) % $p;
// y must be even now
// y = y/2
$y = $y >> 1;
$x = ($x * $x) % $p;
}
return $res;
}
// Returns true if square root
// of n under modulo p exists
// Assumption: p is of the
// form 3*i + 4 where i >= 1
function squareRoot($n, $p)
{
if ($p % 4 != 3)
{
echo "Invalid Input";
return;
}
// Try "+(n^((p + 1)/4))"
$n = $n % $p;
$x = power($n, ($p + 1) / 4, $p);
if (($x * $x) % $p == $n)
{
echo "Square root is ", $x;
return;
}
// Try "-(n ^ ((p + 1)/4))"
$x = $p - $x;
if (($x * $x) % $p == $n)
{
echo "Square root is ", $x;
return;
}
// If none of the above
// two work, then square
// root doesn't exist
echo "Square root doesn't exist ";
}
// Driver Code
$p = 7;
$n = 2;
squareRoot($n, $p);
// This code is contributed by ajit
?>
java 描述语言
<script>
// An efficient Javascript program to find square root under
// modulo p when p is 7, 11, 19, 23, 31, ... etc.
// Utility function to do modular exponentiation.
// It returns (x^y) % p.
function power(x,y,p)
{
let res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y %2== 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns true if square root of n under modulo p exists
// Assumption: p is of the form 3*i + 4 where i >= 1
function squareRoot(n, p)
{
if (p % 4 != 3)
{
document.write("Invalid Input");
return;
}
// Try "+(n^((p + 1)/4))"
n = n % p;
let x = power(n, Math.floor((p + 1) / 4), p);
if ((x * x) % p == n) {
document.write("Square root is " + x);
return;
}
// Try "-(n ^ ((p + 1)/4))"
x = p - x;
if ((x * x) % p == n) {
document.write("Square root is " + x);
return;
}
// If none of the above two work, then
// square root doesn't exist
document.write("Square root doesn't exist ");
}
// Driver program to test
let p = 7;
let n = 2;
squareRoot(n, p);
// This code is contributed by avanitrachhadiya2155
</script>
输出:
Square root is 4
此解决方案的时间复杂度为 O(Log p) 这是如何工作的？ 我们在之前的帖子里已经讨论过欧拉准则。
As per Euler's criterion, if square root exists, then
following condition is true
n(p-1)/2 % p = 1
Multiplying both sides with n, we get
n(p+1)/2 % p = n % p ------ (1)
Let x be the modulo square root. We can write,
(x * x) ≡ n mod p
(x * x) ≡ n(p+1)/2 [Using (1) given above]
(x * x) ≡ n(2i + 2) [Replacing n = 4*i + 3]
x ≡ ±n(i + 1) [Taking Square root of both sides]
x ≡ ±n(p + 1)/4 [Putting 4*i + 3 = p or i = (p-3)/4]
当 p 不是上述形式时，我们将很快讨论方法。 本文由希瓦姆·古普塔供稿。如果发现有不正确的地方，请写评论，或者想分享更多关于以上讨论话题的信息
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