C 解法, 执行用时: 1ms, 内存消耗: 304KB, 提交时间: 2021-05-22

#include <stdio.h>
int main()
{
    double a,b;
    char c;
    while(scanf("%lf%c%lf\n",&a,&c,&b)!=EOF)
    {
        if(c!='+'&&c!='-'&&c!='*'&&c!='/')
        {
            printf("Invalid operation!\n");
        }
        else
        {
            if(c=='/')
            {
                if(b==0)
                {
                    printf("Wrong!Division by zero!\n");
                }
                else
                    printf("%.4lf%c%.4lf=%.4lf\n",a,c,b,a/b);
            }
            else
            {
                if(c=='+')
                    printf("%.4lf%c%.4lf=%.4f\n",a,c,b,a+b);
                else if(c=='-')
                    printf("%.4lf%c%.4lf=%.4lf\n",a,c,b,a-b);
                else
                    printf("%.4lf%c%.4lf=%.4lf\n",a,c,b,a*b);
            }
        }
    }
    return 0;
}

C 解法, 执行用时: 1ms, 内存消耗: 316KB, 提交时间: 2021-11-21

#include<stdio.h>
int main()
{
    double a,b;
    char ch;
    while(scanf("%lf %c %lf",&a,&ch,&b)!=EOF)
    {
        if(ch=='+'||ch=='-'||ch=='*'||ch=='/')
        {
            if(ch=='+')
                printf("%.4lf%c%.4lf=%.4lf\n",a,ch,b,a+b);
                else if(ch=='-')
                    printf("%.4lf%c%.4lf=%.4lf\n",a,ch,b,a-b);
            else if(ch=='*')
                printf("%.4lf%c%.4lf=%.4lf\n",a,ch,b,a*b);
            else
            {
                if(b==0.0)
                    printf("Wrong!Division by zero!\n");
                else
                    printf("%.4lf%c%.4lf=%.4lf\n",a,ch,b,a/b);
            }
        }
        else
            printf("Invalid operation!\n");
    }
    return 0;
}

C 解法, 执行用时: 1ms, 内存消耗: 348KB, 提交时间: 2021-03-03

#include<stdio.h>
int main()
{
    double a,c,d;
    char b;
    while(scanf("%lf%c%lf",&a,&b,&c)!=EOF)
    {
        if(b=='+')
        {
            d=a+c;
            printf("%.4lf+%.4lf=%.4lf\n",a,c,d);
        }
        else if(b=='-')
        {
            d=a-c;
            printf("%.4lf-%.4lf=%.4lf\n",a,c,d);
        }
        else if(b=='*')
        {
            d=a*c;
            printf("%.4lf*%.4lf=%.4lf\n",a,c,d);
        }
        else if(b=='/')
        {
            d=a/c;
            if(c==0.0) printf("Wrong!Division by zero!\n");
            else printf("%.4lf/%.4lf=%.4lf\n",a,c,d);
        }
        else printf("Invalid operation!\n");
    }
    return 0;
}

C 解法, 执行用时: 1ms, 内存消耗: 356KB, 提交时间: 2020-12-20

#include <stdio.h>
int main ()
{
    double n1 = 0.0;
    double n2 = 0.0;
    char op = 0;
    while (~scanf("%lf%c%lf",&n1,&op,&n2))
    {
        switch(op)
        {
            case '+':
                printf("%.4lf+%.4lf=%.4lf\n",n1,n2,n1+n2);
                break;
            case '-':
                printf("%.4lf-%.4lf=%.4lf\n",n1,n2,n1-n2);
                break;
            case '*':
                printf("%.4lf*%.4lf=%.4lf\n",n1,n2,n1*n2);
                break;
            case '/':
                {
                    if(n2==0.0)
                    {
                        printf("Wrong!Division by zero!\n");
                    }
                    else
                    {
                        printf("%.4lf/%.4lf=%.4lf\n",n1,n2,n1/n2);
                    }
                }
               
                break;
            default:
                printf("Invalid operation!\n");
        }
    }
        return 0;
}

C 解法, 执行用时: 1ms, 内存消耗: 364KB, 提交时间: 2021-03-03

#include<stdio.h>
int main()
{
    double a,b;//最初用float，通过率90%
    char o;
    while(scanf("%lf%c%lf",&a,&o,&b)!=EOF)
    {
        switch(o)
        {
            case '+':printf("%.4f+%.4f=%.4f\n",a,b,a+b);break;
            case '-':printf("%.4f-%.4f=%.4f\n",a,b,a-b);break;
            case '*':printf("%.4f*%.4f=%.4f\n",a,b,a*b);break;
            case '/':
            if(b==0)
            {
                printf("Wrong!Division by zero!\n");
            }
            else
            {
                printf("%.4f/%.4f=%.4f\n",a,b,a/b);
            }
            break;
            default:printf("Invalid operation!\n");break;
        }
    }
    return 0;
}