Old Calculator

描述

szhhck have an old calculator bought 5 years ago.he find the old machine can just calculate expressions like this  :

A-B、A+B、A*B、A/B、A%B.

because it is too old and long time not use,the old machine maybe conclude a wrong answer sometime.

Your task is to write a program to check the answer the old calculator calculates is correct or not.

输入

First input is a single line,it's N and stands for there are N test cases.then there are N lines for N cases,each line contain an equation like A op B = C(A,B and C are all integers,and op can only be + , - , * , / or % ).
More details in the Sample Input.

输出

For each test case,if the equation is illegal(divided or
mod by zero),you should Output "Input Error".and if the equation is
correct,Output "Accept";if not Output "Wrong Answer",and print the right
answer after a blank line.

样例输入

5
1+2=32
2-3=-1
4*5=20
6/0=122
8%9=0

样例输出

Wrong Answer
3
Accept
Accept
Input Error
Wrong Answer
8

 #include <stdio.h>

 int main(){
     int T;
     int a;
     int b;
     int c;
     char sign;

     scanf("%d",&T);

     while(T--){
         scanf("%d%c%d=%d",&a,&sign,&b,&c);

         if(sign=='+'){
             if(a+b==c)
                 printf("Accept\n");

             else
                 printf("Wrong Answer\n%d\n",a+b);
         }

         else if(sign=='-'){
             if(a-b==c)
                 printf("Accept\n");

             else
                 printf("Wrong Answer\n%d\n",a-b);
         }

         else if(sign=='*'){
             if(a*b==c)
                 printf("Accept\n");

             else
                 printf("Wrong Answer\n%d\n",a*b);
         }

         else if(sign=='/'){
             if(b==)
                 printf("Input Error\n");

             else if(a/b==c)
                 printf("Accept\n");

             else
                 printf("Wrong Answer\n%d\n",a/b);
         }

         else if(sign=='%'){
             if(b==)
                 printf("Input Error\n");

             else if(a%b==c)
                 printf("Accept\n");

             else
                 printf("Wrong Answer\n%d\n",a%b);
         }
     } 

     return ;
 }