C ++程序查找两个数之和与N都相同的乘积
在本教程中,我们将讨论一个程序来查找两个数字(例如“a”和“b”),以便
a+b = N and a*b = N are satisfied.
从两个方程中消除“a”,我们在“b”和“N”中得到一个二次方程,即
b2 - bN + N = 0
该等式将具有两个根,这将使我们获得“a”和“b”的值。使用行列式方法找到根,我们得出'a'和'b'的值为:
$a=(N-\sqrt{N*N-4N)}/2\\b=(N+\sqrt{N*N-4N)}/2$
示例
#include <iostream> //header file for the square root function #include <math.h> using namespace std; int main() { float N = 12,a,b; cin >> N; //using determinant method to find roots a = (N + sqrt(N*N - 4*N))/2; b = (N - sqrt(N*N - 4*N))/2; cout << "The two integers are :" << endl; cout << "a - " << a << endl; cout << "b - " << b << endl; return 0; }
输出结果
The two integers are : a - 10.899 b - 1.10102