PTA Exercise 2.1 Simple Calculator

题目

模拟简单运算器的工作.假设计算器只能进行加减乘除运算,运算数和结果都是整数,四种运算符的优先级相同,按从左到右的顺序计算.

输入格式:
输入在一行中给出一个四则运算算式,没有空格,且至少有一个操作数.遇等号”=”说明输入结束.

输出格式:
在一行中输出算式的运算结果,或者如果除法分母为0或有非法运算符,则输出错误信息“ERROR”.

输入样例:

1+2*10-10/2=

输出样例:

10

解答

#include<stdio.h>
int main(){
    
    int num,result;
    char ch;
    scanf("%d%c",&num,&ch);
    result=num;
    while(ch!='='){
           
        scanf("%d",&num);
        switch (ch){
    
        case '+':
            result+=num;
            break;
        case '-':
            result-=num;
            break;
        case '*':
            result*=num;
            break;
        case '/':
            if(num==0){
    
                printf("ERROR");
                return 0;
            }
            result/=num;
            break;
        default:
            printf("ERROR");
            return 0;
        }
        
        scanf("%c",&ch);
    }
    printf("%d",result);
    return 0;
}

题目本身不难,But I didn't notice it at first “四种运算符的优先级相同” 这句话,It was a mistake to do a lot of work in vain.

小插曲：
The first name I gave the operator was operator,But this will give an error.It must be a duplicate of the keyword.