给定两个整数n和p，请求出封闭区间[1，n]中x 2 = 1(mod p)的积分解的数量。
例子：
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. 一种简单的解决方案是遍历从1到n的所有数字。对于每个数字，请检查其是否满足方程式。我们可以避免遍及整个范围。该想法基于这样一个事实，即如果数字x满足方程式，则x + i * p形式的所有数字也都满足方程式。我们遍历从1到p的所有数字，对于满足方程的每个数字x，我们发现x + i * p形式的数字计数。为了找到计数，我们首先找到给定x的最大数字，然后将(largest-number – x)/ p加到结果中。
以下是该想法的实现。
C++
// C++ program to count number of values
// that satisfy x^2 = 1 mod p where x lies
// in range [1, n]
#include
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 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
int 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 code
public static void main (String[] args)
{
int n = 10;
int p = 5;
System.out.println(
findCountOfSolutions(n, p));
}
}
// This code is contributed by ajitPython3
# 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 code
n = 10;
p = 5;
print(findCountOfSolutions(n, p));
# This code is contributed by mitsC#
// 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 code
static public void Main ()
{
int n = 10;
int p = 5;
Console.WriteLine(
findCountOfSolutions(n, p));
}
}
// This code is contributed by ajitPHP
$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
输出：
4