YTU-OJ-Problem F: A代码完善--简易二元运算

Problem F: A代码完善--简易二元运算

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 221   Solved: 137
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Description

注：本题只需要提交填写部分的代码，请按照C++方式提交。

编写二元运算类，实现整数的加、减、乘和除四种运算。

#include <stdio.h>
#include <iostream>
using namespace std;
class FourArithOper
{
private:
    int operand1,operand2;
    char operator1;
public:
    FourArithOper(char op,int op1,int op2)
    {
        operand1=op1;
        operand2=op2;
        operator1=op;
    }
    void Algorithm()
    {
        cout<<operand1<<operator1<<operand2<<"=";
/**************************************************   
       请在该部分补充缺少的代码
**************************************************/
        cout<<endl;
    }
};
int  main()
{
    int op1,op2;
    char op;
    cin>>op>>op1>>op2; //操作符，运算数1，运算数2
    FourArithOper fa(op,op1,op2);
    fa.Algorithm();
    return 0;
}

Input

运算符，运算数1，运算数2

Output

按照运算符计算的结果(除法运算保留整数部分)

Sample Input

+ 10 20

Sample Output

10+20=30

HINT

#include <stdio.h>
#include <iostream>
using namespace std;
class FourArithOper
{
private:
    int operand1,operand2;
    char operator1;
public:
    FourArithOper(char op,int op1,int op2)
    {
        operand1=op1;
        operand2=op2;
        operator1=op;
    }
    void Algorithm()
    {
        cout<<operand1<<operator1<<operand2<<"=";
//  请在该部分补充缺少的代码
        if(operator1=='+')
            cout<<operand1+operand2;
        else if(operator1=='-')
            cout<<operand1-operand2;
        else if(operator1=='*')
            cout<<operand1*operand2;
        else if(operator1=='/')
            cout<<operand1/operand2;
//
        cout<<endl;
    }
};
int  main()
{
    int op1,op2;
    char op;
    cin>>op>>op1>>op2; //操作符，运算数1，运算数2
    FourArithOper fa(op,op1,op2);
    fa.Algorithm();
    return 0;
}