HDU1170

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
//HDU1170 基础题//Balloon Comes! http://acm.hdu.edu.cn/showproblem.php?pid=1170//2017.09.15 by wyj#include#includeusing namespace std;struct C {char oper;int x, y;};int main(){int n;double r;scanf("%d", &n);while (n--) {C c;cin >> c.oper >> c.x >> c.y;if (c.oper == '+')printf("%d\n", c.x + c.y);else if (c.oper == '-')printf("%d\n", c.x - c.y);else if (c.oper == '*')printf("%d\n", c.x * c.y);else if (c.oper == '/')c.x % c.y ? printf("%.2lf\n",(double)c.x / c.y) : printf("%d\n", c.x / c.y);}}