ACM HDU p1170 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

这里需要注意的是小数点的取舍   System.out.printf("%.2f",(float)x/y);

package HDU;import java.util.Scanner;public class P1170 {public static void main(String[] args) {     Scanner sc=new Scanner(System.in);     int n=sc.nextInt();     while(n-->0){     String a=sc.next();     char b=a.charAt(0);      int  x= sc.nextInt();      int y=sc.nextInt();      int sum=0;     if(b=='+'){     sum=x+y;     System.out.println(sum);     }     if(b=='-'){     sum=x-y;     System.out.println(sum);     }     if(b=='*'){     sum=x*y;     System.out.println(sum);     }     if(b=='/'){     if(x%y==0){     System.out.println(x/y);          }//这里注意，看清题目意思              else{                System.out.printf("%.2f",(float)x/y);               System.out.println();             }     }          }}}