这两天得知初二的表妹学了一元二次方程,听说还不会解,我就想着试试用C语言编写解方程。
这种方法效果很好:
#include"funct.h"
void yyec1()
{
double a, b, c;
double x = MIN, y;
cout << "-----------" << endl;
cout << "对于ax^2+bx+c=0" << endl;
cout << "依降次输入各项系数 a,b,c" << endl;
cout << "-----------" << endl;
cin >> a >> b >> c;
x = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
y = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
cout << "-----------" << endl;
cout << "x1="<<x << endl <<"x2="<< y << endl;
}
这种好像不是太好:
#include"funct.h"
void yyec2()
{
double a, b, c;
double x = MIN, y;
cout << "-----------" << endl;
cout << "对于ax^2+bx+c=0" << endl;
cout << "依降次输入各项系数 a,b,c" << endl;
cout << "-----------" << endl;
cin >> a >> b >> c;
for (;; x++)
{
y = x / 10;
if (a * y * y + b * y + c == 0)
{
cout << "-----------" << endl;
cout << "x=" << y << endl;
// break;
}
}
}
#include"funct.h"
void yysc()
{
double a, b, c, d;
double x = MIN, y;
cout << "-----------" << endl;
cout << "对于ax^3+bx^2+cx+d=0" << endl;
cout << "依降次输入各项系数 a,b,c,d" << endl;
cout << "-----------" << endl;
cin >> a >> b >> c >> d;
for (;; x++)
{
y = x / 10;
if (a * y * y * y + b * y * y + c * y + d == 0)
{
cout << "-----------" << endl;
cout << "x=" << y << endl;
break;
}
}
}
这个就和二次很像,不做多研究。
#include"funct.h"
void eyyc1()
{
double a, b, c, d, e, f, x, y;
cout << "-----------" << endl;
cout << "对于方程组:" << endl;
cout << "1. ax+by=c " << endl;
cout << "1. dx+ey=f " << endl;
cout << "依次输入a,b,c,d,e,f:" << endl;
cout << "-----------" << endl;
cin >> a >> b >> c >> d >> e >> f;
y = (c * d - a * f) / (b * d - a * e);
x = (b * f - c * e) / (b * d - a * e);
cout << "-----------" << endl;
cout << "y=" << y << endl;
cout << "x=" << x << endl;
}
公式法的错误确实少点,但这似乎不像计算机。
//此方法不好
#include"funct.h"
void eyyc2()
{
double a, b, c, d, e, f;
cout << "-----------" << endl;
cout << "对于方程组:" << endl;
cout << "1. ax+by=c " << endl;
cout << "1. dx+ey=f " << endl;
cout << "依次输入a,b,c,d,e,f:" << endl;
cout << "-----------" << endl;
cin >> a >> b >> c >> d >> e >> f;
double x = MIN, y = MIN, xx, yy;
if(d!=0){
for (;; y++)
{
yy = y / 100;
// cout<<yy<<endl;
if (yy*b*d-a*e*yy==c*d-a*f)
{
cout << "-----------" << endl;
cout << "y=" << yy << endl;
cout << "x=" << (f-e*yy)/d << endl;
break;
}
// else cout<<"0";
}}
else {
for (;; y++)
{
yy = y / 100;
// cout<<yy<<endl;
if (yy*e==f)
{
cout << "y=" << yy << endl;
cout << "x=" << (c-b*yy)/a << endl;
break;
}
// else cout<<"0";
}
}
}
#include"funct.h"
int main()
{
while (1)
{
cout << "-----------" << endl;
cout << "1.一元二次" << endl;
cout << "2.一元三次" << endl;
cout << "3.二元一次" << endl;
cout << "-----------" << endl;
int se;
cin >> se;
switch(se)
{
case 1:
yyec1();
//yyec2();
break;
case 2:
yysc();
break;
case 3:
eyyc1();
//eyyc2();
break;
default:cout << "what are you inputting?" << endl; break;
}
}
return 0;
}
#include<iostream>
#include <cmath>
#define MIN -100
using namespace std;
void yyec1();
void yyec2();
void yysc();
void eyyc1();
void eyyc2();