Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5402 Accepted Submission(s): 1722
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
package ACM1;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.Scanner;
public class hdu1 {
public static void main(String[]args)
{
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
for(int i=0;i<n;i++)
{
String a = scanner.next();
int b = scanner.nextInt();
int c = scanner.nextInt();
//double result = (double)b/(double)c;
if(a.charAt(0)=='+')
System.out.println((b+c));
else if(a.charAt(0)=='-')
System.out.println((b-c));
else if(a.charAt(0)=='*')
System.out.println((b*c));
else
{
if(b%c==0)
System.out.println((b/c));
else
{
DecimalFormat dc = new DecimalFormat("0.00");
//dc.setMaximumFractionDigits(2);
//dc.setGroupingSize(0);
//dc.setRoundingMode(RoundingMode.FLOOR);
//System.out.println(dc.format(b/(1.0*c)));//我不明白为什么这个样子不对
System.out.format("%.2f",(b/(1.0*c)) ).println();//为什么这个样子对
}
}
}
}
}
