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
- 我的github
我的github:先来贴个图~ 这是我的github,新建了第一个repository,默认路径是aokoqingiz/code. 然后是里面的文件~ 里面有一个readme.txt,是我对这个r ...
- Spring properties dependency checking
In Spring,you can use dependency checking feature to make sure the required properties have been set ...
- SaltStack安装Redis模块
安装redis Python Client 下载地址: https://pypi.python.org/simple/redis/ tar -xvf redis-2.8.0.tar.gz cd red ...
- 怎样提交FIREDAC数据集的DELTA到中间件然后保存进数据库
你可以在客户端序列FireDAC数据集的DELTA , 将序列后的STREAM发送给中间件, 中间件的TFDQuery或TFDMemTable调用LOADFROMSTREAM()方法加载流, 然后调用 ...
- HDU 5754 Life Winner Bo (找规律and博弈)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5754 给你四种棋子,棋子一开始在(1,1)点,两个人B和G轮流按每种棋子的规则挪动棋子,棋子只能往右下 ...
- dll开发中遇到的问题
刚碰到个问题,我的一个项目中引用了一个dll,这个dll又引用了另一个dll,我把这俩个都放在bin文件夹下,但是会报错,说第二个dll找不到.把它放到系统文件夹system32下就没事了. 但是遇到 ...
- JS escape、encodeURI 、encodeURIComponent 编码与解码[转]
转至:http://jc-dreaming.iteye.com/blog/1702407 本文讨论如何对传递参数用JS编码与解码 1:编码与解码方法的对应关系 escape ------------- ...
- 一款多浏览器兼容的javascript多级下拉菜单
这个多级下拉菜单的脚本大小不到2K,带有动画效果,可以方便地支持多个实例,并且能良好兼容WordPress系统wp_list_cats和wp_list_pages生成的多级列表.要初始化一个菜单,只需 ...
- map的正确删除方式
遍历删除map元素的正确方式是 for(itor = maptemplate.begin; itor != maptemplate.end(); ) { if(neederase) ...
- SQLite使用事务更新—by command
public void SaveToDB(DataTable dt) { /* todo:sqlite没有提供批量插入的机制,需要通过事务处理 更新所有数据 * http://www.cnblogs. ...