NYOJ 486 Old Calculator

比眉伴天荒 2022-09-19 08:08 174阅读 0赞

## 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

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

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;
    		}
    	}
    }