求方程ax²+bx+c=0的根
我的代碼
//求方程ax²+bx+c=0的根
#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;
double fun(int a, int b, int c)
{
double temp, x, x1 = 0, x2 = 0;
//double temp;
if(a == 0)
if(b == 0)
{
cout << "此方程無意義" << endl;
}
else
{
x = -c/b;
cout << "x = " << x << endl;
}
else
{
if(b*b - 4*a*c < 0)
{
cout << "此方程無實根" << endl;
//return 0;
}
else
{
temp = sqrt(double(b*b - 4*a*c));
x1 = (-b + temp)/(2*a);
x2 = (-b - temp)/(2*a);
cout << "x1 = " << x1 << "," << "x2 = " << x2 << endl;
//return 0;
}
}
return 0;
}
int main()
{
cout << "請輸入二元一次方程參數a、b、c:";
int a, b, c;
cin >> a >> b >> c;
//cout << fun(a,b,c) << endl;
fun(a,b,c);
system("pause");
return 0;
}
參考代碼
百度文庫 50道C++編程練習題
總結
判斷二元一次方程有無實根條件:a ≠ 0;b² - 4ac ≥0;
sqrt函數在math.h中,應用時須在表達式後加float,double,或 long double才能編譯通過,例如 a = sqrt(float(n));
返回值return加在函數最後就可以,條件判斷末尾不用加;
函數fun中已經用cout輸出方程的解,主函數直接調用就可以打印出解(cout << fun(a,b,c) << endl;畫蛇添足,連返回值0都打印出來)