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
3
12
-12
1200
Sample Output
21
-21
2100
注意：前导0的情况！
例：
输入：
3
-0012560020
00000
00205
输出为：
-2006521
0
502
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 instr = Integer.parseInt(str); //System.out.println(instr); str = Integer.toString(instr); //System.out.println(str); if (str.charAt(0) == '-') { System.out.print("-"); int k = 0; boolean isOne=false; //System.out.println(str.length()+"aaa"); for (int i = str.length() - 1; i >= 1; i--) { //System.out.println("a: "+str.charAt(i)); if(str.charAt(i)!='0'&&!isOne){ //System.out.println("++ "+str.charAt(i)); isOne=true; } if (isOne) { System.out.print(str.charAt(i)); k++; } } for (int i = 1; i < str.length() - k; i++) { System.out.print(0); } System.out.println(); } else { int k = 0; boolean isOne=false; for (int i = str.length() - 1; i >= 0; i--) { if(str.charAt(i)!='0'&&!isOne){ isOne=true; } if (isOne) { System.out.print(str.charAt(i)); k++; } } for (int i = 0; i < str.length() - k; i++) { System.out.print(0); } System.out.println(); } } } }
