Count number of solutions of x^2 = 1 (mod p) in given range
Given two integers n and p, find the number of integral solutions to x2 = 1 (mod p) in the closed interval [1, n].
Examples:
Input : n = 10, p = 5 Output : 4 There are four integers that satisfy the equation x2 = 1. The numbers are 1, 4, 6 and 9. Input : n = 15, p = 7 Output : 5 There are five integers that satisfy the equation x2 = 1. The numbers are 1, 8, 15, 6 and 13.
One simple solution is to go through all numbers from 1 to n. For every number, check if it satisfies the equation. We can avoid going through the whole range. The idea is based on the fact that if a number x satisfies the equation, then all numbers of the form x + i*p also satisfy the equation. We traverse for all numbers from 1 to p and for every number x that satisfies the equation, we find the count of numbers of the form x + i*p. To find the count, we first find the largest number for given x and then add (largest-number – x)/p to the result.
Below is the implementation of the idea.
C++
// C++ program to count number of values// that satisfy x^2 = 1 mod p where x lies// in range [1, n]#include<bits/stdc++.h>using namespace std;typedef long long ll;int findCountOfSolutions(int n, int p){ // Initialize result ll ans = 0; // Traverse all numbers smaller than // given number p. Note that we don't // traverse from 1 to n, but 1 to p for (ll x=1; x<p; x++) { // If x is a solution, then count all // numbers of the form x + i*p such // that x + i*p is in range [1,n] if ((x*x)%p == 1) { // The largest number in the // form of x + p*i in range // [1, n] ll last = x + p * (n/p); if (last > n) last -= p; // Add count of numbers of the form // x + p*i. 1 is added for x itself. ans += ((last-x)/p + 1); } } return ans;}// Driver codeint main(){ ll n = 10, p = 5; printf("%lld\n", findCountOfSolutions(n, p)); return 0;} |
Java
// Java program to count// number of values that// satisfy x^2 = 1 mod p// where x lies in range [1, n]import java.io.*;class GFG{static int findCountOfSolutions(int n, int p){ // Initialize result int ans = 0; // Traverse all numbers // smaller than given // number p. Note that // we don't traverse from // 1 to n, but 1 to p for (int x = 1; x < p; x++) { // If x is a solution, // then count all numbers // of the form x + i*p // such that x + i*p is // in range [1,n] if ((x * x) % p == 1) { // The largest number // in the form of x + // p*i in range [1, n] int last = x + p * (n / p); if (last > n) last -= p; // Add count of numbers // of the form x + p*i. // 1 is added for x itself. ans += ((last - x) / p + 1); } } return ans;}// Driver codepublic static void main (String[] args){ int n = 10; int p = 5; System.out.println( findCountOfSolutions(n, p));}}// This code is contributed by ajit |
Python3
# Program to count number of# values that satisfy x^2 = 1# mod p where x lies in range [1, n]def findCountOfSolutions(n, p): # Initialize result ans = 0; # Traverse all numbers smaller # than given number p. Note # that we don't traverse from # 1 to n, but 1 to p for x in range(1, p): # If x is a solution, then # count all numbers of the # form x + i*p such that # x + i*p is in range [1,n] if ((x * x) % p == 1): # The largest number in the # form of x + p*i in range # [1, n] last = x + p * (n / p); if (last > n): last -= p; # Add count of numbers of # the form x + p*i. 1 is # added for x itself. ans += ((last - x) / p + 1); return int(ans);# Driver coden = 10;p = 5;print(findCountOfSolutions(n, p)); # This code is contributed by mits |
C#
// C# program to count// number of values that// satisfy x^2 = 1 mod p// where x lies in range [1, n]using System;class GFG{static int findCountOfSolutions(int n, int p){ // Initialize result int ans = 0; // Traverse all numbers // smaller than given // number p. Note that // we don't traverse from // 1 to n, but 1 to p for (int x = 1; x < p; x++) { // If x is a solution, // then count all numbers // of the form x + i*p // such that x + i*p is // in range [1,n] if ((x * x) % p == 1) { // The largest number // in the form of x + // p*i in range [1, n] int last = x + p * (n / p); if (last > n) last -= p; // Add count of numbers // of the form x + p*i. // 1 is added for x itself. ans += ((last - x) / p + 1); } } return ans;}// Driver codestatic public void Main (){ int n = 10; int p = 5; Console.WriteLine( findCountOfSolutions(n, p));}}// This code is contributed by ajit |
PHP
<?php// Program to count number of// values that satisfy x^2 = 1// mod p where x lies in range [1, n]function findCountOfSolutions($n, $p){ // Initialize result $ans = 0; // Traverse all numbers smaller // than given number p. Note // that we don't traverse from // 1 to n, but 1 to p for ($x = 1; $x < $p; $x++) { // If x is a solution, then // count all numbers of the // form x + i*p such that // x + i*p is in range [1,n] if (($x * $x) % $p == 1) { // The largest number in the // form of x + p*i in range // [1, n] $last = $x + $p * ($n / $p); if ($last > $n) $last -= $p; // Add count of numbers of // the form x + p*i. 1 is // added for x itself. $ans += (($last - $x) / $p + 1); } } return $ans;}// Driver code$n = 10;$p = 5;echo findCountOfSolutions($n, $p); // This code is contributed by ajit?> |
Javascript
<script>// Javascript program to count number// of values that satisfy x^2 = 1 mod p// where x lies in range [1, n]function findCountOfSolutions(n, p){ // Initialize result let ans = 0; // Traverse all numbers smaller // than given number p. Note that // we don't traverse from 1 to n, // but 1 to p for(let x = 1; x < p; x++) { // If x is a solution, // then count all numbers // of the form x + i*p // such that x + i*p is // in range [1,n] if ((x * x) % p == 1) { // The largest number // in the form of x + // p*i in range [1, n] let last = x + p * (n / p); if (last > n) last -= p; // Add count of numbers // of the form x + p*i. // 1 is added for x itself. ans += ((last - x) / p + 1); } } return ans;} // Driver codelet n = 10;let p = 5;document.write(findCountOfSolutions(n, p)); // This code is contributed by susmitakundugoaldanga </script> |
Output:
4
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