HDU 1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17032    Accepted Submission(s): 6226

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

Ｃ表示运算　Ａ,Ｂ为操作数 求输出的结果．．如果是整数直接输出，浮点数输出两位小数

#include<iostream>#include<cstdio>using namespace std;int main(){    int a,b;    int n;    char c;    while(cin>>n)    {      while(n--)      {          cin>>c>>a>>b;        switch(c)        {           case'+':cout<<a+b<<endl;break;           case'-':cout<<a-b<<endl;break;           case'*':cout<<a*b<<endl;break;           case'/':if(a%b==0) cout<<a/b<<endl;          else printf("%.2f\n",1.0*a/b);break;        }      }    }    return 0;}