题解 | #简单计算器#

#include <cstdlib>
#include <iomanip>
#include <iostream>
#include <string>
using namespace std;

int main() {
    int i=0, b=0,n=0;
    string calc;
    string num1,num2,op;
    getline(cin, calc);//直接cin出现空格会中断，使用getline函数，跳过空格中断，将空格计入字符串
    while (calc[i]!='\0') {
        if(calc[i]=='+')//判断是不是合法的运算符号，并用b作为标志位记录相应的运算
        {
            b=1;
            n=i;
        }
        else if(calc[i]=='-')
        {
            b=2;
            n=i;
        }
        else if(calc[i]=='*')
        {
            b=3;
            n=i;
        }
        else if(calc[i]=='/')
        {
            b=4;
            n=i;
        }
        else if((calc[i]<48||calc[i]>57)&&calc[i]!='.')
        {
            b=0;
            n=i;
        }
        i++;
    }
    num1 = calc.substr(0,n);//将输入的字符串里数字的部分分割出来
    int len=calc.length();//重点在于确定运算符的位置便于进行分割
    num2 = calc.substr(n+1,len+1);//裁剪第二个运算数字
    double number1=stod(num1);//将字符串转换double类型的数据
    double number2=stod(num2);
    if(number2-0.0<1e-10)//对第二位数进行判断，若为0则按非法输入判断
    {
        b=5;
    }
    
    switch (b) {
        case 0:cout<<"Invalid operation!";break;
        case 1:cout<<setprecision(4)<<fixed<<number1<<"+"<<number2<<"="<<number1+number2;break;
        case 2:cout<<setprecision(4)<<fixed<<number1<<"-"<<number2<<"="<<number1-number2;break;
        case 3:cout<<setprecision(4)<<fixed<<number1<<"*"<<number2<<"="<<number1*number2;break;
        case 4:cout<<setprecision(4)<<fixed<<number1<<"/"<<number2<<"="<<number1/number2;break;
        case 5:cout<<"Wrong!Division by zero!";break;
    
    }

    
}
// 64 位输出请用 printf("%lld")