NYOJ 486 Old Calculator

Old Calculator

时间限制：1000 ms  |  内存限制：65535 KB

难度：1

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

51+2=322-3=-14*5=206/0=1228%9=0

样例输出

Wrong Answer3AcceptAcceptInput ErrorWrong Answer8

思路：略

 

#include<stdio.h>int main(){int m;scanf("%d",&m);while(m--){int a,b,sum;char ch;scanf("%d%c%d=%d",&a,&ch,&b,&sum);switch(ch){case '+':if (a+b == sum){printf("Accept\n");}else{printf("Wrong Answer\n%d\n",a+b);}break;case '-':if (a-b == sum){printf("Accept\n");}else{printf("Wrong Answer\n%d\n",a-b);}break;case '*':if (a*b == sum){printf("Accept\n");}else{printf("Wrong Answer\n%d\n",a*b);}break;case '/':if (b == 0){printf("Input Error\n");}else{if (a/b == sum){printf("Accept\n");}else{printf("Wrong Answer\n%d\n",a/b);}}break;case '%':if (b == 0){printf("Input Error\n");}else{if (a%b == sum){printf("Accept\n");}else{printf("Wrong Answer\n%d\n",a%b);}}break;}}}