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
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