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
代码语言：javascript复制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();
}
}
}
}
