C++一题请教高手帮忙解决
答案:5 悬赏:40 手机版
解决时间 2021-04-16 19:23
- 提问者网友：心牵心
- 2021-04-15 22:15
C++一题请教高手帮忙解决
最佳答案
- 五星知识达人网友：三千妖杀
- 2021-04-15 22:47
以下是完整代码。。参考一下。。
#include<iostream>
#include<cmath>
using namespace std;
void fun(double a, double b, double c)
{
double d = b*b - 4*a*c;
double x1,x2;
if (d < 0)
{
cout <<"方程无实根\n";
}
else if (d == 0.0)
{
cout <<"方程有两个相同的实根：\n";
x1 = -b/(2*a);
cout << "x1=x2=" << x1 << endl;
}
else if (d > 0)
{
cout <<"方程有两个不同的实根：\n";
x1 = (-b+sqrt(d))/(2*a);
x2 = (-b-sqrt(d))/(2*a);
cout << "x1=" << x1 << "，x2=" << x2 << endl;
}
}
void main()
{
double a, b, c;
cout << "请输入一元二次方程的三个系数：\n";
cout <<"a=";
cin >> a;
cout <<"b=";
cin >> b;
cout <<"c=";
cin >> c;
cout << endl;
fun(a,b,c);
}
#include<iostream>
#include<cmath>
using namespace std;
void fun(double a, double b, double c)
{
double d = b*b - 4*a*c;
double x1,x2;
if (d < 0)
{
cout <<"方程无实根\n";
}
else if (d == 0.0)
{
cout <<"方程有两个相同的实根：\n";
x1 = -b/(2*a);
cout << "x1=x2=" << x1 << endl;
}
else if (d > 0)
{
cout <<"方程有两个不同的实根：\n";
x1 = (-b+sqrt(d))/(2*a);
x2 = (-b-sqrt(d))/(2*a);
cout << "x1=" << x1 << "，x2=" << x2 << endl;
}
}
void main()
{
double a, b, c;
cout << "请输入一元二次方程的三个系数：\n";
cout <<"a=";
cin >> a;
cout <<"b=";
cin >> b;
cout <<"c=";
cin >> c;
cout << endl;
fun(a,b,c);
}
全部回答
- 1楼网友：毛毛
- 2021-04-16 01:58
#include "stdafx.h"
#include "iostream.h"
#include "math.h"
void fun(double a,double b,double c)
{
double x1=(-b+sqrt(pow(b,2)-4*a*c))/2*a;
double x2=(-b-sqrt(pow(b,2)-4*a*c))/2*a;
cout<<"x1:"<<x1<<endl;
cout<<"x2:"<<x2<<endl;
}
int main(int argc, char* argv[])
{
double a,b,c;
int i=0;
while(pow(b,2)<4*a*c||i==0)
{
if(i) cout<<"error! enter again!"<<endl;
cout<<"enter a:"<<endl;
cin>>a;
cout<<"enter b:"<<endl;
cin>>b;
cout<<"enter c:"<<endl;
cin>>c;
++i;
}
fun(a,b,c);
return 0;
}
- 2楼网友：长青诗
- 2021-04-16 01:41
#include <math.h>
#include <stdio.h>
void fun( double a, double b, double c )
{
double delta = b*b-4*a*c;
double x1, x2;
if ( delta < 0 ) {
printf( "\n方程无实根" );
} else {
x1 = ( -b + sqrt( delta ) ) / 2*a;
x2 = ( -b - sqrt( delta ) ) / 2*a;
if ( delta != 0 ) {
printf( "\n方程有两个不同的实根:\n\n" );
printf( "x1=%.2f x2=%.2f\n\n", x1, x2 );
} else {
printf( "\n方程有两个相同的实根:\n\n" );
printf( "x1=x2=%.2f\n\n", x1 );
}
}
}
int main()
{
double a, b, c;
printf( "请输入一元二次方程的三个系数：\n" );
scanf( "a=%lf, b=%lf, c=%lf", &a, &b, &c );
fun( a, b, c );
}
- 3楼网友：持酒劝斜阳
- 2021-04-16 00:33
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
int main()
{
void fun(double a,double b,double c);
double a,b,c;
cout<<"请输入a，b，c的值："<<endl;
cin>>a>>b>>c;
fun(a,b,c);
system("PAUSE");
return 0;
}
void fun(double a,double b,double c)
{
float disc,x1,x2,p,q;
disc=pow(b,2)-4*a*c;
if(disc>=0)
{
p=-b/(2*a); q=sqrt(disc)/(2*a);
x1=p+q; x2=p-q;
cout<<"x1="<<x1<<" "<<"x2="<<x2;
}
else
cout<<"Error";
}
- 4楼网友：思契十里
- 2021-04-15 23:41
#include<iostream.h> #include<math.h> void fun(double a, double b, double c) { double disc,x1,x2,p,q; if(fabs(a)<1e-6) cout<<"不是一元二次方程!"<<endl; else { disc=b*b-4*a*c; if(fabs(disc)<=1e-6) //判断是否有两个相等的实根 cout<<"x1-x2="<<-b/(2*a)<<endl; else { if(disc>1e-6) //判断有两个不相等的实根 { x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); cout<<"x1="<<x1<<endl; cout<<"x2="<<x2<<endl; } else { p=-b/(2*a); q=sqrt(fabs(disc))/(2*a); cout<<"x1="<<p<<"+"<<q<<"i"<<endl; cout<<"x2="<<p<<"-"<<q<<"i"<<endl; } } }
} void main() { double a,b,c; cout<<"请输入3个系数:"<<endl; cin>>a>>b>>c; fun(a,b,c); } //注意我们的disc定义成的是个实数所以不能直接的和0进行比较。1e-6表示的是一个很小的数。用这种方法来判断disc和0的关系
我要举报
如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
点此我要举报以上问答信息
大家都在看
推荐资讯