满足给定方程的整数对(a，b)
给定方程组 a 2 + b = n 和 a + b 2 = m 。任务是找出满足给定的 n 和 m 等式的正整数对 (a，b) 的个数。 举例:
输入: n = 9，m = 3 输出: 1 解释: 满足条件的两个方程只存在一对(3，0)。 输入: n = 4，m = 20 输出: 0 说明: 不存在这样的对。
方法: 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。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)中获得 a 的值。
- 如果配对 (a，b) 满足方程(1)，那么配对 (a，b) 就是方程组的解。
以下是上述方法的实现:
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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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
</script>
Output:
1
时间复杂度: O(sqrt(min(n，m))
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