满足给定方程的一对整数(a，b)

给定方程组a ² + b = n和a + b ² = m 。任务是找到满足给定n和m方程的正整数对(a，b)的数量。
例子：

Input: n = 9, m = 3 
Output: 1 
Explanation: 
Only one pair (3, 0) exists for both equations satisfying the conditions.
Input: n = 4, m = 20 
Output: 0 
Explanation: 
There are no such pair exists. 
 

方法：
该方法是检查所有可能的数字对，并检查该对是否满足两个方程式。为此，我们有

a2 + b = n ... (1)
   a + b2 = m ... (2)
For equation (2), 
=> a = m - b2 ... (3)

• 现在，对于a的正值，b的每个值都必须从0到sqrt(m)。
• 从等式(3)获得a的值。
• 如果对(a，b)满足方程式(1)，则对(a，b)是方程组的解。

下面是上述方法的实现：

C++

// C++ program to count the pair of integers(a, b)
// which satisfy the equation
// a^2 + b = n and a + b^2 = m
#include 
using namespace std;
 
// Function to count valid pairs
int pairCount(int n, int m)
{
    int cnt = 0, b, a;
    for (b = 0; b <= sqrt(m); b++) {
        a = m - b * b;
        if (a * a + b == n) {
            cnt++;
        }
    }
    return cnt;
}
 
// Driver code
int main()
{
    int n = 9, m = 3;
 
    cout << pairCount(n, m) << endl;
 
    return 0;
}

Java

// Java program to count the pair of integers(a, b)
// which satisfy the equation
// a^2 + b = n and a + b^2 = m
class GFG
{
 
// Function to count valid pairs
static int pairCount(int n, int m)
{
    int cnt = 0, b, a;
    for (b = 0; b <= Math.sqrt(m); b++)
    {
        a = m - b * b;
        if (a * a + b == n)
        {
            cnt++;
        }
    }
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 9, m = 3;
    System.out.print(pairCount(n, m) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count the pair of integers(a, b)
# which satisfy the equation
# a^2 + b = n and a + b^2 = m
 
# Function to count valid pairs
def pairCount(n, m):
    cnt = 0;
    for b in range(int(pow(m, 1/2))):
        a = m - b * b;
        if (a * a + b == n):
            cnt += 1;
         
    return cnt;
 
# Driver code
if __name__ == '__main__':
    n = 9;
    m = 3;
    print(pairCount(n, m));
     
# This code is contributed by 29AjayKumar

C#

// C# program to count the pair of integers(a, b)
// which satisfy the equation
// a^2 + b = n and a + b^2 = m
using System;
 
class GFG
{
 
// Function to count valid pairs
static int pairCount(int n, int m)
{
    int cnt = 0, b, a;
    for (b = 0; b <= Math.Sqrt(m); b++)
    {
        a = m - b * b;
        if (a * a + b == n)
        {
            cnt++;
        }
    }
    return cnt;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 9, m = 3;
    Console.Write(pairCount(n, m) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program to count the pair of integers(a, b)
// which satisfy the equation
// a^2 + b = n and a + b^2 = m
 
// Function to count valid pairs
function pairCount(n , m)
{
    var cnt = 0, b, a;
    for (b = 0; b <= Math.sqrt(m); b++)
    {
        a = m - b * b;
        if (a * a + b == n)
        {
            cnt++;
        }
    }
    return cnt;
}
 
// Driver code
var n = 9, m = 3;
document.write(pairCount(n, m));
 
// This code is contributed by Amit Katiyar

输出：

1

时间复杂度： O(sqrt(min(n，m))