您的位置:首页 > 移动开发 > IOS开发

第九周实验报告5

2012-04-17 18:39 387 查看
#include "iostream"
using namespace std;
class CEquation
{
private:
double a;     // 未知数系数
double b;     // 常数项
char unknown; // 未知数的符号
public:
CEquation(double aa=0,double bb=0);
friend istream &operator >> (istream &in,CEquation &e);
friend ostream &operator << (ostream &out,CEquation &e);
double Solve();
char getUnknown();
};

CEquation::CEquation(double aa,double bb):a(aa),b(bb){}

// 输入方程
istream &operator >> (istream &in,CEquation &e)
{
char ch1,ch2,ch3,ch4;
while(1)
{
cin>>e.a>>ch1>>ch2>>e.b>>ch3>>ch4;
if (ch1>='a' && ch1<='z')
if ((ch2=='+' || ch2=='-') && ch3=='=' && ch4=='0') break;
cout<<"输入的方程格式不符合规范,请重新输入\n";
}
if (ch2=='-') e.b=-e.b;
e.unknown=ch1;
return in;
}

// 输出方程
ostream &operator << (ostream &out,CEquation &e)
{
cout<<e.a<<e.unknown;
if (e.b>=0) cout<<"+";
cout<<e.b<<"=0"<<endl;
return out;
}

// 求解
double CEquation::Solve()
{
double x;
if (a==0)
{
if (b==0) cout<<"任意一个实数均为方程的解。"<<endl;
else cout<<"方程无解。"<<endl;
return 0;
}
x=-b/a;
return x;
}

char CEquation::getUnknown()
{
return unknown;
}

int main()
{
CEquation e;
cout<<"请输入一元一次方程(输入格式:3x-8=0):";
cin>>e;   //在两次测试中,分别输入3x-8=0和5s+18=0
cout<<"方程为:"<<e;
cout<<"方程的解为:"<<e.getUnknown()<<"="<<e.Solve()<<endl; //对两次测试,分别输出x=...和s=...
e.Solve();
system("pause");
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息