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 } } } } }
时间: 2024-10-14 18:24:15