Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18296    Accepted Submission(s): 6753

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

3-120.50

 

Author

lcy

 

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#include<iostream>#include<cstdio>using namespace std;int main(){    int i,j,t,a,b;    char c;    scanf("%d",&t);     while(t--)     {       getchar();       scanf("%c %d %d",&c,&a,&b);       //cin>>c>>a>>b;       //printf("fkjkf");       if(c=='-')       {           printf("%d\n",a-b);       }       else       if(c=='+')       {           printf("%d\n",a+b);       }       else       if(c=='*')       {           printf("%d\n",a*b);       }       else       if(c=='/')       {         double e;         e=a/(b*1.0);         if(e==(int)(e))         printf("%d\n",(int)e);         else         printf("%.2f\n",e);       }      /* if(c=='/')       {           if(a%b!=0)           {               printf("%.2f\n",(double)a/b);           }           else           printf("%d\n",a/b);       }*/    }    return 0;}