Problem Description

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6452 Accepted Submission(s): 2004 

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

// Balloon Comes!# include<stdio.h>int main(){  int n;  char a[10];  while(scanf("%d",&n))  {  while(n--)  {  int b,c;  int i=0;  scanf("%s%d%d",a,&b,&c);  //对于char类型，在输入时有一个空格，因此需要声明一个char组来吸收空格   if(a[0]=='+')  printf("%d\n",b+c);  else if(a[0]=='-')  printf("%d\n",b-c);  else if(a[0]=='*')  printf("%d\n",b*c);  else if(a[0]=='/')  {  if(b%c==0)      //注意能除尽与不能除尽的情况   printf("%d\n",b/c);  else printf("%.2f\n",(float)b/c);}  }  }return 0;}