HDOJ1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32182    Accepted Submission(s): 12105

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 -1 2 0.50

基础题

import java.util.Scanner;

public class Main{
  private static Scanner scanner;
  public static void main(String[] args) {
    scanner = new Scanner(System.in);
    int cases = scanner.nextInt();
    while(cases-->0){
      String string = scanner.next();
      int a = scanner.nextInt();
      int b = scanner.nextInt();
      if(string.equals("+")){
        System.out.println(a+b);
      }else if(string.equals("-")){
        System.out.println(a-b);
      }else if(string.equals("*")){
        System.out.println(a*b);
      }else {
        double res = 1.0*a/b;
        double r = res-(int)res;
        if(r>0){
          System.out.printf("%.2f",res);
        }else {
          System.out.print((int)res);
        }
        System.err.println();
      }
    }
  }
}