Hdu1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28744 Accepted Submission(s): 10830

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

题意：输入t表示有t组数据，然后就是两个数字的运算。

心得：本来不想把这道题写到这里来的，因为水，但做着做着发现了一些知识上的漏洞，所以决定把写下来。

            这道题，我是用Java写的，第一次提交WA，以为是要hasNext()，改了以后提交，TLE，又测试了一下，发现除得到整数的情况有问题，原来Java中的Double型都是带小数点的，1就是1.0，这点原来还真没注意，于是判断了一下，如果结果是整数就用int相除，否则用结果为double型。也可以先取余判断是否能除尽。

import java.util.Scanner;public class Main {public static void main(String[] args) {         Scanner sc = new Scanner(System.in);         int t = sc.nextInt();                  while(t-->0){         String str = sc.next();         int a = sc.nextInt();         int b = sc.nextInt();         double sum=0;         if(str.equals("+")){         System.out.println(a+b);         }else if(str.equals("-")){         System.out.println(a-b);         }else if(str.equals("*")){         System.out.println(a*b);         }else if(str.equals("/")){         sum=a*1.0/b;         int sum2 = a/b;         if(sum2==sum){         System.out.println(sum2);         }else{         System.out.printf("%.2f",sum);         System.out.println();         }         }                  } }}