Problem description

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.

解题思路:先把两个字符串中的大写字母全部换成小写字母,然后用strcmp(s1,s2)函数比较两个字符串的大小,水过。

AC代码:

 #include<bits/stdc++.h>
using namespace std;
int main(){
char s1[],s2[];
cin>>s1>>s2;
for(int i=;s1[i]!='\0';++i){
if(s1[i]>='A'&&s1[i]<='Z')s1[i]+=;
if(s2[i]>='A'&&s2[i]<='Z')s2[i]+=;
}
if(strcmp(s1,s2)>)cout<<""<<endl;
else if(strcmp(s1,s2)<)cout<<"-1"<<endl;
else cout<<""<<endl;
return ;
}

A - Petya and Strings的更多相关文章

  1. codeforces水题100道 第二十四题 Codeforces Beta Round #85 (Div. 2 Only) A. Petya and Strings (strings)

    题目链接:http://www.codeforces.com/problemset/problem/112/A题意:忽略大小写,比较两个字符串字典序大小.C++代码: #include <cst ...

  2. codeforces 112APetya and Strings(字符串水题)

    A. Petya and Strings 点击打开题目 time limit per test 2 seconds memory limit per test 256 megabytes input ...

  3. Codeforces 112A-Petya and Strings(实现)

    A. Petya and Strings time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  4. Codeforces 149 E. Martian Strings

    正反两遍扩展KMP,维护公共长度为L时.出如今最左边和最右边的位置. . .. 然后枚举推断... E. Martian Strings time limit per test 2 seconds m ...

  5. CodeForces832-B. Petya and Exam

    补的若干年以前的题目,水题,太菜啦_(:з」∠)_    B. Petya and Exam time limit per test 2 seconds memory limit per test 2 ...

  6. Codeforces Round #425 (Div. 2) Problem B Petya and Exam (Codeforces 832B) - 暴力

    It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy ...

  7. Codeforces Round #425 (Div. 2) B. Petya and Exam(字符串模拟 水)

    题目链接:http://codeforces.com/contest/832/problem/B B. Petya and Exam time limit per test 2 seconds mem ...

  8. CodeForces 832B Petya and Exam

    B. Petya and Exam time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  9. Codeforces Round #425 (Div. 2) B - Petya and Exam

    地址:http://codeforces.com/contest/832/problem/B 题目: B. Petya and Exam time limit per test 2 seconds m ...

随机推荐

  1. json 添加 和删除两种方法

    <script> var test = { name: "name", age: "12" }; var countrys = { "ne ...

  2. 一次vue-cli 2.x项目打包优化经历(优化xlsx插件)

    一.分析各模块打包后大小 用vue-cli创建的项目,已经集成 webpack-bundle-analyzer.详见文件 build/webpack.prod.conf.js,代码如下: if (co ...

  3. vue监听数组中某个属性,计算其他属性问题

    今天在项目开发中遇到一个根据数组中某个属性变化同时更新另一个属性变化的问题,刚开始代码如下 this.weekList1=r.data.roomProducts; this.weekList1.map ...

  4. Spring学习总结(4)——Spring AOP教程

    一.概念 AOP(Aspect Oriented Programming):面向切面编程. 面向切面编程(也叫面向方面编程),是目前软件开发中的一个热点,也是Spring框架中的一个重要内容.利用AO ...

  5. MINSUB - Largest Submatrix

    MINSUB - Largest Submatrix no tags  You are given an matrix M (consisting of nonnegative integers) a ...

  6. sql server的数据库个数、表个数及表的数据量统计

    sql server的数据库个数.表个数及表的数据量统计   --由于今天要监控数据,急需统计实例中1有多少库2库里有多少表3每个表有多少数据 --将写好的代码贴出来,用到如下的: --sysobje ...

  7. nyoj 309

    #include<stdio.h> #include<stdlib.h> #define N 1100 int c[N]; int main() { int l,n,i,m,a ...

  8. Blue Jeans POJ 3080 寻找多个串的最长相同子串

    Description The Genographic Project is a research partnership between IBM and The National Geographi ...

  9. HTML5:判断浏览器是否支持date类型

    在某些情况下,我们需要判断当前浏览器是否支持date类型,如果支持的话,可以让用户用原生的datepicker来选取日期.如果不支持,则我们需要用自己实现的datepicker类库,来提供给用户. 在 ...

  10. 【原创】TCP超时重传机制探索

    TCP超时重传机制探索 作者:tll (360电商技术) 1)通信模型 TCP(Transmission Control Protocol)是一种可靠传输协议.在传输过程中当发送方(sender)向接 ...