hdu 1266 Reverse Number
2015-07-20 21:58
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题意:将一个正整数颠倒,如后输出这个数。
注意:整数中间的0是合法的。
Total Submission(s): 6078 Accepted Submission(s): 2774
Problem Description
Welcome to 2006'4 computer college programming contest!
欢迎来到2006年4月大学生计算机程序设计竞赛!
Specially, I give my best regards to all freshmen!
特别的,我给每一位大一新生给予真诚的致谢!
You are the future of HDU ACM!
也许你将改变HDU ACM的未来!
And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
现在,我必须告诉你 HDU ACM的题目并不简单,但是,这题一定简单。。。哈哈!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
给你一个整数,你的任务是颠倒这个数字,然后输出。颠倒规则如下:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
一般颠倒方式:把一个尾部没有0的正整数颠倒,例如12颠倒后就是21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
负数颠倒:例如-12颠倒后就是-21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
尾部有0:例如1200颠倒2100。
Input
Input file contains multiple test cases.
输入文件将包含多组测试事件。
There is a positive integer n (n<100) in the first line, which means the number of test cases,
输入的第一行是一个正整数,小于100,表示测试事件的个数,
and then n 32-bit integers follow.
然后如如32位地整数(c语言中的int)。
Output
For each test case, you should output its reverse number, one case per line.
对于每一个测试事件,你应该输出颠倒后的结果,一行一个答案。
Sample Input
Sample Output
Author
lcy
注意:整数中间的0是合法的。
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); while (n-- > 0) { String s = sc.next(); String result = ""; char c; int index = s.length() - 1;// 扩大作用域 for (int i = index; i > 0; i--) {// 第一个不扫描 c = s.charAt(i);// c只是一个零时变量,方便编程 if (c != '0') { index = i; break; } } for (int i = index; i > 0; i--) {// 第一个不扫描 result += s.charAt(i); } c = s.charAt(0); if (c == '-') { result = "-" + result;// 负数符号放前面 } else { result += c;// 整数放后面 } // 后面加0 for (int j = index; j < s.length() - 1; j++) { result += '0'; } System.out.println(result); } } }
Reverse Number
颠倒数字
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6078 Accepted Submission(s): 2774
Problem Description
Welcome to 2006'4 computer college programming contest!
欢迎来到2006年4月大学生计算机程序设计竞赛!
Specially, I give my best regards to all freshmen!
特别的,我给每一位大一新生给予真诚的致谢!
You are the future of HDU ACM!
也许你将改变HDU ACM的未来!
And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
现在,我必须告诉你 HDU ACM的题目并不简单,但是,这题一定简单。。。哈哈!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
给你一个整数,你的任务是颠倒这个数字,然后输出。颠倒规则如下:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
一般颠倒方式:把一个尾部没有0的正整数颠倒,例如12颠倒后就是21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
负数颠倒:例如-12颠倒后就是-21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
尾部有0:例如1200颠倒2100。
Input
Input file contains multiple test cases.
输入文件将包含多组测试事件。
There is a positive integer n (n<100) in the first line, which means the number of test cases,
输入的第一行是一个正整数,小于100,表示测试事件的个数,
and then n 32-bit integers follow.
然后如如32位地整数(c语言中的int)。
Output
For each test case, you should output its reverse number, one case per line.
对于每一个测试事件,你应该输出颠倒后的结果,一行一个答案。
Sample Input
3 12 -12 1200
Sample Output
21 -21 2100
Author
lcy
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