Codeforces刷题之路——112A Petya and Strings

A. Petya and Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.

Input

Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.

Output

If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.

Examples

input

aaaa
aaaA

output

0

input

abs
Abz

output

-1

input

abcdefg
AbCdEfF

output

1

Note

If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:

• http://en.wikipedia.org/wiki/Lexicographical_order

题目大意：比较字符串，不考虑大小写，即A与a同为一个字母，第一个出现字母不相同的字符进行比较，相等输出0，小于输出-1，大于输出1。

解题思路：将其全部转换成小写或者大写，循环寻找第一个不相同的字符，进行大小比较即可。

以下为解题代码（java实现）

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		String str1 = scanner.nextLine().toLowerCase();
		String str2 = scanner.nextLine().toLowerCase();
		
		boolean flag = false;
		char[] ch1 = str1.toCharArray();
		char[] ch2 = str2.toCharArray();
		
		for(int i = 0;i < ch1.length;i++){
			if(ch1[i] == ch2[i]){
				flag = true;
			}else{
				flag = false;
				if(ch1[i] < ch2[i]){
						System.out.println(-1);
				}else{
					System.out.println(1);
				}
				break;
			}
		}
		if(flag){
			System.out.println(0);
		}
		scanner.close();
	}
}