CF Tanya and Postcard
2 seconds
256 megabytes
standard input
standard output
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s.
The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS".
Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message.
The first line contains line s (1 ≤ |s| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), consisting of uppercase and lowercase English letters — the text written in the newspaper.
Here |a| means the length of the string a.
Print two integers separated by a space:
- the first number is the number of times Tanya shouts "YAY!" while making the message,
- the second number is the number of times Tanya says "WHOOPS" while making the message.
AbC
DCbA
3 0
ABC
abc
0 3
abacaba
AbaCaBA
3 4
题目太难懂,三个人读出三种意思。
#include <iostream>
#include <cctype>
#include <cstring>
#include <cstdio>
using namespace std; int T_L_HAVE[] = {};
int S_L_HAVE[] = {};
int T_U_HAVE[] = {};
int S_U_HAVE[] = {}; int main(void)
{
string s;
string t;
int len_s,len_t;
int sum_yay = ;
int sum_whoops = ; cin >> s >> t;
len_s = s.size();
len_t = t.size(); for(int i = ;i < len_t;i ++)
if(islower(t[i]))
T_L_HAVE[t[i] - 'a'] ++;
else
T_U_HAVE[t[i] - 'A'] ++; for(int i = ;i < len_s;i ++)
if(islower(s[i]))
S_L_HAVE[s[i] - 'a'] ++;
else
S_U_HAVE[s[i] - 'A'] ++; for(int i = ;i < len_s;i ++)
{
if(islower(s[i]) && T_L_HAVE[s[i] - 'a'])
{
S_L_HAVE[s[i] - 'a'] --;
T_L_HAVE[s[i] - 'a'] --;
sum_yay ++;
}
else if(isupper(s[i]) && T_U_HAVE[s[i] - 'A'])
{
S_U_HAVE[s[i] - 'A'] --;
T_U_HAVE[s[i] - 'A'] --;
sum_yay ++;
}
} for(int i = ;i < len_s;i ++)
{
if(islower(s[i]) && S_L_HAVE[s[i] - 'a'] && T_U_HAVE[s[i] - - 'A'])
{
S_L_HAVE[s[i] - 'a'] --;
T_U_HAVE[s[i] - - 'A'] --;
sum_whoops ++;
}
else if(isupper(s[i]) && S_U_HAVE[s[i] - 'A'] && T_L_HAVE[s[i] + - 'a'])
{
S_U_HAVE[s[i] - 'A'] --;
T_L_HAVE[s[i] + - 'a'] --;
sum_whoops ++;
}
} cout << sum_yay << ' ' << sum_whoops << endl; return ;
}
CF Tanya and Postcard的更多相关文章
- CodeForces 518B. Tanya and Postcard
B. Tanya and Postcard time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- Codeforces Round #293 (Div. 2) B. Tanya and Postcard 水题
B. Tanya and Postcard time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- codeforces 518B. Tanya and Postcard 解题报告
题目链接:http://codeforces.com/problemset/problem/518/B 题目意思:给出字符串 s 和 t,如果 t 中有跟 s 完全相同的字母,数量等于或者多过 s,就 ...
- CodeForces 518B Tanya and Postcard (题意,水题)
题意:给定两个字符串,然后从第二个中找和第一个相同的,如果大小写相同,那么就是YAY,如果大小写不同,那就是WHOOPS.YAY要尽量多,其次WHOOPS也要尽量多. 析:这个题并不难,难在读题懂题意 ...
- Codeforces Round #293 (Div. 2)
A. Vitaly and Strings 题意:两个字符串s,t,是否存在满足:s < r < t 的r字符串 字符转处理:字典序排序 很巧妙的方法,因为s < t,只要找比t字典 ...
- codeforces518B
Tanya and Postcard CodeForces - 518B 有个小女孩决定给他的爸爸寄明信片.她已经想好了一句话(即长度为n的字符串s),包括大写和小写英文字母.但是他不会写字,所以她决 ...
- CF 1005A Tanya and Stairways 【STL】
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairw ...
- CF 584B Kolya and Tanya
题目大意:3n个人围着一张桌子,给每个人发钱,可以使1块.2块.3块,第i个人的金额为Ai.若存在第个人使得Ai + Ai+n + Ai+2n != 6,则该分配方案满足条件,求所有的满足条件的方案数 ...
- CF 508D Tanya and Password(无向图+输出欧拉路)
( ̄▽ ̄)" //不知道为什么,用scanf输入char数组的话,字符获取失效 //于是改用cin>>string,就可以了 //这题字符的处理比较麻烦,输入之后转成数字,用到函 ...
随机推荐
- ProtoBuffer 简单例子
最近学了一下protobuf,写了一个简单的例子,如下: person.proto文件 message Person{ required string name = 1; required int32 ...
- MVC架构和SSH框架对应关系
MVC三层架构:模型层(model),控制层(controller)和视图层(view).模型层,用Hibernate框架让来JavaBean在数据库生成表及关联,通过对JavaBean的操作来对数据 ...
- [iOS微博项目 - 1.7] - 版本新特性
A.版本新特性 1.需求 第一次使用新版本的时候,不直接进入app,而是展示新特性界面 github: https://github.com/hellovoidworld/HVWWeibo ...
- 启动报错:java.lang.ClassNotFoundException: org.springframework.web.context.ContextLoaderListener
如果你是maven项目,tomcat在发布项目的时候没有同时发布maven依赖所添加的jar包,你需要设置一下eclipse:项目 —> 属性 -> Deployment Assembly ...
- jquery ajax请求后台 的简单例子
jQuery.ajax(url,[settings]) 概述 通过 HTTP 请求加载远程数据. jQuery 底层 AJAX 实现.简单易用的高层实现见 $.get, $.post 等.$.ajax ...
- Unity3D音乐音效学习笔记
对于Unity3D的音乐音效这块一直没有好好的看过,现在准备好好的研究一下,并作为一个笔记记录下. 支持格式 在游戏中,一般存在两种音乐,一种是时间较长的背景音乐,一种是时间较短的音效(比如按钮点击, ...
- Linux rsync 同步实践
目录[-] 1. rsync 同步的大致思路 2. rsync的安装 3. rsync的配置 4. rsync的基本操作 服务器端启动 注2. 实时同步 注3. rsync通过linux防火墙 公司网 ...
- java spring 使用注解来实现缓存
这里举例使用spring3.1.4 + ehcache 注解的方式使用cache 是在spring3.1加入的 使用方法: 1.ehcache依赖+spring依赖 <!-- ehcache依赖 ...
- _vsnprintf 用法
_vsnprintf,C语言库函数之一,属于可变参数.用于向字符串中打印数据.数据格式用户自定义. 头文件: #include <stdarg.h> 函数声明: int _vsnprint ...
- mysql 重要维护工具 图解
下载地址: http://maatkit.org/get/mk-query-digest更多信息: http://maatkit.org/ | http://code.google.com/p ...