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Balloon Comes!

2024-09-27 21:01:39   216人阅读

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23182    Accepted Submission(s): 8774


Problem Description
The contest starts now! How excited it is to see balloons floating around. You, one of the best programmers in HDU, can get a very beautiful balloon if only you have solved the very very very... easy problem.
Give you an operator (+,-,*, / --denoting addition, subtraction, multiplication, division respectively) and two positive integers, your task is to output the result.
Is it very easy?
Come on, guy! PLMM will send you a beautiful Balloon right now!
Good Luck!
 


 

Input
Input contains multiple test cases. The first line of the input is a single integer T (0<T<1000) which is the number of test cases. T test cases follow. Each test case contains a char C (+,-,*, /) and two integers A and B(0<A,B<10000).Of course, we all know that A and B are operands and C is an operator.
 


 

Output
For each case, print the operation result. The result should be rounded to 2 decimal places If and only if it is not an integer.
 


 

Sample Input
4
+ 1 2
- 1 2
* 1 2
/ 1 2
 


 

Sample Output
3
-1
2
0.50
 


 

 

 

#include<stdio.h>
int main()
{
 int  n,a,b;
 char k;
 scanf("%d",&n);
 while(n--)
 {
    getchar();
      scanf("%c %d%d",&k,&a,&b);
    if(k==‘+‘)
        printf("%d\n",a+b);
    if(k==‘-‘)
       printf("%d\n",a-b);
     if(k==‘*‘)
      printf("%d\n",a*b);
    if(k==‘/‘)
    {
      if(a%b==0)
          printf("%d\n",a/b);
      else 
          printf("%.2f\n",a*1.0/b);
    } 
 }
 return 0;
}


 

Balloon Comes!


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