C++新手题，谢谢帮忙
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悬赏园豆：20
[已解决问题]
解决于 2020-11-30 22:52
Activity 4
Write a program to calculate the roots of a quadratic equation. A quadratic equation is an equation of the form:
ax^2+bx+c=0
The roots of a quadratic equation are given by the formula:
x=(-b±√(b^2-4ac))/2a
Your program should do the following:
- Ask the user to enter values for a, b, and c. For example, print “a = ” to the console and then read the value for a.
- Determine if the equation has 0, 1, or 2 roots. Print a message display the number of roots, for example: “The equation has 2 roots” or “The equation has 1 root” (Technically 1 solution is called a double root, but we’ll just call it 1 root for this activity).
- Calculate and print the roots (if any). For a single root, print “r = <VALUE>”. For 2 roots, print “r1 = <VALUE>” on one line and “r2 = <VALUE>” on the next.
活动4
编写一个程序来计算二次方程的根。二次方程式为以下形式的方程式：
ax ^ 2 + bx + c = 0
二次方程的根由下式给出：
x =（-b±√（b ^ 2-4ac））/ 2a
您的程序应执行以下操作：
1.要求用户输入a，b和c的值。例如，在控制台上打印“ a =”，然后读取a的值。
2.确定方程式是否具有0、1、2的根。打印一条消息，显示根数，例如：“等式具有2个根”或“等式具有1个根”（技术上1个解称为双根，但在此活动中我们仅称其为1个根） 。
3.计算并打印根（如果有）。对于单个根，打印“ r = <VALUE>”。对于2个根，在一行上打印“ r1 = <VALUE>”，在下一行上打印“ r2 = <VALUE>”。
最佳答案
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写好了，不过内容输出使用的是英文，注释上有中文，如果你不是小白的话，应该懂得怎么替换的吧。
代码放上来了，如果哪看不懂的话，我还可以写注释，或者私聊更详细的讲。
#include <iostream> #include <cmath> using namespace std; void print(double a,double b,double c); void count(double a,double b,double c); int main() { double a,b,c; cout<<"Please input the a,b,c of the equation:"<<endl; // cout<<"请输入方程的a，b，c："<<endl; cout<<"\ta = "; cin>>a; cout<<"\tb = "; cin>>b; cout<<"\tc = "; cin>>c; cout<<"Yours input is:"<<endl; // cout<<"你的输入是："<<endl; print(a,b,c); count(a,b,c); return 0; } void print(double a,double b,double c) { cout<<'\t'; if(a!=0) { if(a==1) cout<<"x^2"; else cout<<a<<"x^2"; } if(b!=0) { if(b==1) cout<<"+x"; else if(b==-1) cout<<"-x"; else if(b>0) cout<<"+"<<b<<"x"; else if(b<0) cout<<b<<"x"; } if(c!=0) { if(c>0) cout<<"+"; cout<<c; } cout<<"=0"<<endl; } void count(double a,double b,double c) { double dierta,r,r1,r2; dierta=b*b-4*a*c; if(dierta<0) cout<<"Equation has no root."<<endl; // cout<<"等式没有根"<<endl; else if(dierta==0) { r=-b/(2.0*a); cout<<"Equation has one root:"<<endl; // cout<<"等式具有1个根"<<endl; cout<<"\tr = "<<r<<endl; } else if(dierta>0) { r1=(-b+sqrt(dierta))/(2.0*a); r2=(-b-sqrt(dierta))/(2.0*a); cout<<"Equation has two roots:"<<endl; // cout<<"等式具有2个根"<<endl; cout<<"\tr1 = "<<r1<<endl; cout<<"\tr2 = "<<r2<<endl; } }
收获园豆：20
可以添加个好友么？我的题有点多。希望大神能够帮忙。
@心律: 可不是免费劳动的噢
@Conan-jine: 已经发布了 还是20 。谢谢啦。
