HACM 1170 Balloon Comes!

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

 

#include <iostream> #include<string> using namespace std; int main() { int n; char s; int a, b; while (cin>>n) { for (int i = 0; i < n; i++) { cin >> s >> a >> b; getchar(); if (s== '+') cout<<a+b<<endl; else if (s == '-') cout<<a-b<<endl; else if (s== '*') cout<<a*b<<endl; else  if (a%b != 0)//a 不可以整除b，则输出保留2位有效数字 { cout.precision(2); cout << fixed <<1.0*a/b<<endl; } else cout<<a / b<<endl; } } return 0; }