HDOJ1266 Reverse Number

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10371    Accepted Submission(s): 4625

Problem Description

Welcome to 2006'4 computer college programming contest!

Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!

Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

 

Input

Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

 

Output

For each test case, you should output its reverse number, one case per line.

 

Sample Input

312-121200

 

Sample Output

21-212100

 一个个加到string 里面，后面的0统计一下，最后加上。

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) {int n = scanner.nextInt();if (n == 0) {System.out.println(0);continue;}String string = "";if (n < 0) {string += "-";n = -n;}int count = 0;boolean boo = true;while (n > 0) {if (n % 10 != 0 || !boo) {string += ("" + n % 10);boo = false;} else if (n % 10 == 0 && boo) {count++;}n /= 10;}for (int i = 0; i < count; i++) {string += ("" + 0);}System.out.println(string);}}}