给定两个数字 a 和 b，求所有 x，使得 a % x = b
原文:https://www . geesforgeks . org/given-two-numbers-b-find-x-x-b/
给定两个数字 a 和 b，求所有 x，使得 a % x = b。
示例:
Input : a = 21, b = 5
Output : 2
The answers of the Modular Equation are
8 and 16 since 21 % 8 = 21 % 16 = 5 .
这里出现 3 种情况:
- If (a < b) then there is no answer.
- If (a = b) then all numbers greater than a are the answers, then there will be infinite possible solutions.
- If (a > b) let x be the answer to our equation. Then divide x by (a–b). Similarly, because a% x = b, then b < x. These conditions are also necessary and sufficient. Therefore, the answer is that the divisor of A–B is strictly greater than B can be solved in O (SQRT (a-b)) . There is only one case here. When (a-b) is a perfect square, we have to deal with it separately, and then we add its square root twice, so we have to subtract it once. If this happens.
c++
// C++ program to find x such that a % x is equal
// to b.
#include <bits/stdc++.h>
using namespace std;
void modularEquation(int a, int b)
{
// if a is less than b then no solution
if (a < b) {
cout << "No solution possible " << endl;
return;
}
// if a is equal to b then every number
// greater than a will be the solution
// so its infinity
if (a == b) {
cout << "Infinite Solution possible " << endl;
return;
}
// all resultant number should be greater than
// b and (a-b) should be divisible by resultant
// number
// count variable store the number of values
// possible
int count = 0;
int n = a - b;
int y = sqrt(a - b);
for (int i = 1; i <= y; ++i) {
if (n % i == 0) {
// checking for both divisor and quotient
// whether they divide ( a-b ) completely
// and greater than b .
if (n / i > b)
count++;
if (i > b)
count++;
}
}
// Here y is added twice in the last iteration
// so 1 y should be decremented to get correct
// solution
if (y * y == n && y > b)
count--;
cout << count << endl;
}
// Driver code
int main()
{
int a = 21, b = 5;
modularEquation(a, b);
return 0;
}
Java
// Java program to find x such that
// a % x is equal to b.
import java.io.*;
class GFG {
static void modularEquation(int a, int b)
{
// if a is less than b then no solution
if (a < b) {
System.out.println("No solution possible ");
return;
}
// if a is equal to b then every number
// greater than a will be the solution
// so its infinity
if (a == b) {
System.out.println("Infinite Solution possible ");
return;
}
// all resultant number should be greater
// than b and (a-b) should be divisible
// by resultant number
// count variable store the number of
// values possible
int count = 0;
int n = a - b;
int y = (int)Math.sqrt(a - b);
for (int i = 1; i <= y; ++i) {
if (n % i == 0) {
// checking for both divisor and
// quotient whether they divide
// ( a-b ) completely and
// greater than b .
if (n / i > b)
count++;
if (i > b)
count++;
}
}
// Here y is added twice in the last
// iteration so 1 y should be decremented
// to get correct solution
if (y * y == n && y > b)
count--;
System.out.println(count);
}
// Driver code
public static void main(String[] args)
{
int a = 21, b = 5;
modularEquation(a, b);
}
}
// This code is contributed by Prerna Saini
python 3
# Python3 program to find x such
# that a % x is equal to b.
import math
def modularEquation(a, b) :
# if a is less than b then no solution
if (a < b) :
print("No solution possible ")
return
# if a is equal to b then every number
# greater than a will be the solution
# so its infinity
if (a == b) :
print("Infinite Solution possible ")
return
# all resultant number should be
# greater than b and (a-b) should
# be divisible by resultant number
# count variable store the number
# of values possible
count = 0
n = a - b
y = (int)(math.sqrt(a - b))
for i in range(1, y+1) :
if (n % i == 0) :
# checking for both divisor
# and quotient whether they
# divide ( a-b ) completely
# and greater than b .
if (n / i > b) :
count = count + 1
if (i > b) :
count = count + 1
# Here y is added twice in the
# last iteration so 1 y should be
# decremented to get correct
# solution
if (y * y == n and y > b) :
count = count - 1
print (count)
# Driver code
a = 21
b = 5
modularEquation(a, b)
# This code is contributed by Nikita Tiwari.
T35】c#T3T39】PHPT4T43】JavascriptT46
输出:
2
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