Balloon Comes!

Time Limit: 1000ms   Memory limit: 65535K  有疑问？点这里^_^

题目描述

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!

输入

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.

输出

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.

示例输入

4

+ 1 2

- 1 2

* 1 2

/ 1 2

示例输出

3

-1

2

0.50

#include int main() {     int t, a, b;     char c;     scanf("%d", &t);  getchar();     while (t--) {         scanf("%c", &c);         scanf("%d %d", &a, &b);         switch (c) {             case '+':                 printf("%d", a+b);                 break;             case '-':                 printf("%d", a-b);                 break;             case '*':                 printf("%d", a*b);                 break;             case '/':                 if (a % b == 0)                     printf("%d", a/b);                 else                     printf("%.2lf", (double)a/b);                 break;         }         printf("\n");         getchar();     }     return 0; }