给定A和B，任务是找到X可以采用的可能值的数量，以使给定的模块化方程(A mod X)= B成立。在此，X也称为模块化方程式的解。
例子：
Input : A = 26, B = 2
Output : 6
Explanation
X can be equal to any of {3, 4, 6, 8,
12, 24} as A modulus any of these values
equals 2 i. e., (26 mod 3) = (26 mod 4)
= (26 mod 6) = (26 mod 8) = .... = 2
Input : 21 5
Output : 2
Explanation
X can be equal to any of {8, 16} as A modulus
any of these values equals 5 i.e. (21 mod
8) = (21 mod 16) = 5如果我们仔细分析方程式A mod X = B，则很容易注意到，如果(A = B)，那么X可以取无穷多个比A大的值。在(A
现在，在这种情况下，我们可以使用一个众所周知的关系，即
Dividend = Divisor * Quotient + Remainder我们正在寻找所有可能的X即除数，给定A即股息，而B即余数。所以，
We can say,
A = X * Quotient + B
Let Quotient be represented as Y
∴ A = X * Y + B
A - B = X * Y
∴ To get integral values of Y,
we need to take all X such that X divides (A - B)
∴ X is a divisor of (A - B)因此，问题减少到找到(A – B)的除数，并且这样的除数的数量就是X可以取的可能值。
但是，我们知道A mod X会产生从(0到X – 1)的值，我们必须取所有这样的X使得X>B。
因此，我们可以得出结论：(A – B)的除数大于B是X满足A mod X = B所需的所有可能值。
C++
/* C++ Program to find number of possible
values of X to satisfy A mod X = B */
#include
using namespace std;
/* Returns the number of divisors of (A - B)
greater than B */
int calculateDivisors(int A, int B)
{
int N = (A - B);
int noOfDivisors = 0;
for (int i = 1; i <= sqrt(N); i++) {
// if N is divisible by i
if ((N % i) == 0) {
// count only the divisors greater than B
if (i > B)
noOfDivisors++;
// checking if a divisor isnt counted twice
if ((N / i) != i && (N / i) > B)
noOfDivisors++;
}
}
return noOfDivisors;
}
/* Utility function to calculate number of all
possible values of X for which the modular
equation holds true */
int numberOfPossibleWaysUtil(int A, int B)
{
/* if A = B there are infinitely many solutions
to equation or we say X can take infinitely
many values > A. We return -1 in this case */
if (A == B)
return -1;
/* if A < B, there are no possible values of
X satisfying the equation */
if (A < B)
return 0;
/* the last case is when A > B, here we calculate
the number of divisors of (A - B), which are
greater than B */
int noOfDivisors = 0;
noOfDivisors = calculateDivisors(A, B);
return noOfDivisors;
}
/* Wrapper function for numberOfPossibleWaysUtil() */
void numberOfPossibleWays(int A, int B)
{
int noOfSolutions = numberOfPossibleWaysUtil(A, B);
// if infinitely many solutions available
if (noOfSolutions == -1) {
cout << "For A = " << A << " and B = " << B
<< ", X can take Infinitely many values"
" greater than " << A << "\n";
}
else {
cout << "For A = " << A << " and B = " << B
<< ", X can take " << noOfSolutions
<< " values\n";
}
}
// Driver code
int main()
{
int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21, B = 5;
numberOfPossibleWays(A, B);
return 0;
} Java
/* Java Program to find number of possible
values of X to satisfy A mod X = B */
import java.lang.*;
class GFG
{
/* Returns the number of divisors of (A - B)
greater than B */
public static int calculateDivisors(int A, int B)
{
int N = (A - B);
int noOfDivisors = 0;
for (int i = 1; i <= Math.sqrt(N); i++)
{
// if N is divisible by i
if ((N % i) == 0)
{
// count only the divisors greater than B
if (i > B)
noOfDivisors++;
// checking if a divisor isnt counted twice
if ((N / i) != i && (N / i) > B)
noOfDivisors++;
}
}
return noOfDivisors;
}
/* Utility function to calculate number of all
possible values of X for which the modular
equation holds true */
public static int numberOfPossibleWaysUtil(int A, int B)
{
/* if A = B there are infinitely many solutions
to equation or we say X can take infinitely
many values > A. We return -1 in this case */
if (A == B)
return -1;
/* if A < B, there are no possible values of
X satisfying the equation */
if (A < B)
return 0;
/* the last case is when A > B, here we calculate
the number of divisors of (A - B), which are
greater than B */
int noOfDivisors = 0;
noOfDivisors = calculateDivisors(A, B);
return noOfDivisors;
}
/* Wrapper function for numberOfPossibleWaysUtil() */
public static void numberOfPossibleWays(int A, int B)
{
int noOfSolutions = numberOfPossibleWaysUtil(A, B);
// if infinitely many solutions available
if (noOfSolutions == -1)
{
System.out.print("For A = " + A + " and B = " + B
+ ", X can take Infinitely many values"
+ " greater than " + A + "\n");
}
else
{
System.out.print("For A = " + A + " and B = " + B
+ ", X can take " + noOfSolutions
+ " values\n");
}
}
// Driver program
public static void main(String[] args)
{
int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21;
B = 5;
numberOfPossibleWays(A, B);
}
}
// Contributed by _omgPython3
# Python Program to find number of possible
# values of X to satisfy A mod X = B
import math
# Returns the number of divisors of (A - B)
# greater than B
def calculateDivisors (A, B):
N = A - B
noOfDivisors = 0
a = math.sqrt(N)
for i in range(1, int(a + 1)):
# if N is divisible by i
if ((N % i == 0)):
# count only the divisors greater than B
if (i > B):
noOfDivisors +=1
# checking if a divisor isnt counted twice
if ((N / i) != i and (N / i) > B):
noOfDivisors += 1;
return noOfDivisors
# Utility function to calculate number of all
# possible values of X for which the modular
# equation holds true
def numberOfPossibleWaysUtil (A, B):
# if A = B there are infinitely many solutions
# to equation or we say X can take infinitely
# many values > A. We return -1 in this case
if (A == B):
return -1
# if A < B, there are no possible values of
# X satisfying the equation
if (A < B):
return 0
# the last case is when A > B, here we calculate
# the number of divisors of (A - B), which are
# greater than B
noOfDivisors = 0
noOfDivisors = calculateDivisors;
return noOfDivisors
# Wrapper function for numberOfPossibleWaysUtil()
def numberOfPossibleWays(A, B):
noOfSolutions = numberOfPossibleWaysUtil(A, B)
#if infinitely many solutions available
if (noOfSolutions == -1):
print ("For A = " , A , " and B = " , B
, ", X can take Infinitely many values"
, " greater than " , A)
else:
print ("For A = " , A , " and B = " , B
, ", X can take " , noOfSolutions
, " values")
# main()
A = 26
B = 2
numberOfPossibleWays(A, B)
A = 21
B = 5
numberOfPossibleWays(A, B)
# Contributed by _omgC#
/* C# Program to find number of possible
values of X to satisfy A mod X = B */
using System;
class GFG
{
/* Returns the number of divisors of (A - B)
greater than B */
static int calculateDivisors(int A, int B)
{
int N = (A - B);
int noOfDivisors = 0;
double a = Math.Sqrt(N);
for (int i = 1; i <= (int)(a); i++)
{
// if N is divisible by i
if ((N % i) == 0)
{
// count only the divisors greater than B
if (i > B)
noOfDivisors++;
// checking if a divisor isnt counted twice
if ((N / i) != i && (N / i) > B)
noOfDivisors++;
}
}
return noOfDivisors;
}
/* Utility function to calculate number of all
possible values of X for which the modular
equation holds true */
static int numberOfPossibleWaysUtil(int A, int B)
{
/* if A = B there are infinitely many solutions
to equation or we say X can take infinitely
many values > A. We return -1 in this case */
if (A == B)
return -1;
/* if A < B, there are no possible values of
X satisfying the equation */
if (A < B)
return 0;
/* the last case is when A > B, here we calculate
the number of divisors of (A - B), which are
greater than B */
int noOfDivisors = 0;
noOfDivisors = calculateDivisors(A, B);
return noOfDivisors;
}
/* Wrapper function for numberOfPossibleWaysUtil() */
public static void numberOfPossibleWays(int A, int B)
{
int noOfSolutions = numberOfPossibleWaysUtil(A, B);
// if infinitely many solutions available
if (noOfSolutions == -1)
{
Console.Write ("For A = " + A + " and B = " + B
+ ", X can take Infinitely many values"
+ " greater than " + A + "\n");
}
else
{
Console.Write ("For A = " + A + " and B = " + B
+ ", X can take " + noOfSolutions
+ " values\n");
}
}
public static void Main()
{
int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21;
B = 5;
numberOfPossibleWays(A, B);
}
}
// Contributed by _omgPHP
$B)
$noOfDivisors++;
// checking if a divisor isnt counted twice
if (($N / $i) != $i && ($N / $i) > $B)
$noOfDivisors++;
}
}
return $noOfDivisors;
}
/* Utility function to calculate number of all
possible values of X for which the modular
equation holds true */
function numberOfPossibleWaysUtil($A, $B)
{
/* if A = B there are infinitely many solutions
to equation or we say X can take infinitely
many values > A. We return -1 in this case */
if ($A == $B)
return -1;
/* if A < B, there are no possible values of
X satisfying the equation */
if ($A < $B)
return 0;
/* the last case is when A > B, here we calculate
the number of divisors of (A - B), which are
greater than B */
$noOfDivisors = 0;
$noOfDivisors = calculateDivisors($A, $B);
return $noOfDivisors;
}
/* Wrapper function for numberOfPossibleWaysUtil() */
function numberOfPossibleWays($A, $B)
{
$noOfSolutions = numberOfPossibleWaysUtil($A, $B);
// if infinitely many solutions available
if ($noOfSolutions == -1) {
echo "For A = " , $A, " and B = " ,$B,
"X can take Infinitely many values
greater than " , $A , "\n";
}
else {
echo "For A = ", $A , " and B = " ,$B,
" X can take ",$noOfSolutions,
" values\n";
}
}
// Driver code
$A = 26; $B = 2;
numberOfPossibleWays($A, $B);
$A = 21; $B = 5;
numberOfPossibleWays($A, $B);
// This code is contributed ajit.
?>Javascript
输出：
For A = 26 and B = 2, X can take 6 values
For A = 21 and B = 5, X can take 2 values上述方法的时间复杂度不过是找到(A – B)除数的时间复杂度，即O(√(A – B))