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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
51+2=322-3=-14*5=206/0=1228%9=0
- 样例输出
Wrong Answer3AcceptAcceptInput ErrorWrong Answer8
1 #include <stdio.h> 2 3 int main(){ 4 int T; 5 int a; 6 int b; 7 int c; 8 char sign; 9 10 scanf("%d",&T);11 12 while(T--){13 scanf("%d%c%d=%d",&a,&sign,&b,&c);14 15 if(sign==‘+‘){16 if(a+b==c)17 printf("Accept\n");18 19 else20 printf("Wrong Answer\n%d\n",a+b);21 }22 23 else if(sign==‘-‘){24 if(a-b==c)25 printf("Accept\n");26 27 else28 printf("Wrong Answer\n%d\n",a-b);29 }30 31 else if(sign==‘*‘){32 if(a*b==c)33 printf("Accept\n");34 35 else36 printf("Wrong Answer\n%d\n",a*b);37 }38 39 else if(sign==‘/‘){40 if(b==0)41 printf("Input Error\n");42 43 else if(a/b==c)44 printf("Accept\n");45 46 else47 printf("Wrong Answer\n%d\n",a/b); 48 }49 50 else if(sign==‘%‘){51 if(b==0)52 printf("Input Error\n");53 54 else if(a%b==c)55 printf("Accept\n");56 57 else58 printf("Wrong Answer\n%d\n",a%b); 59 }60 } 61 62 return 0;63 }
Old Calculator
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