CodeForces 518B. 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 这道题目的意思很好理解。
第一种情况,你要的字符串和报纸上的字符串有多少相同的,每个字母相同的个数累加起来再输出。
第二种情况,你要的字符串和报纸上的字符串有多少值相同,但大小写不同,将每个字母的情况累加起来再输出。 在样例里有这样的情况
ZZZzzz
ZZZZzz 答案是:5 1 也就是说,在你统计完第一种情况后,一样的字母需要你减去,然后才能去统计第二种情况。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ctype.h>
#include <queue>
#define mem(a, b) memset((a), (b), (sizeof(a)))
using namespace std;
const int max_size = + ;
char str1[max_size];
char str2[max_size];
int cnt1[];
int cnt2[]; int main()
{
gets(str1);
gets(str2);
int len1 = strlen(str1);
int len2 = strlen(str2);
mem(cnt1, );
mem(cnt2, ); for(int i = ; i < len1; i++)
{
cnt1[str1[i] - 'A']++;
}
for(int i = ; i < len2; i++)
{
cnt2[str2[i] - 'A']++;
} int ans1, ans2;
ans1 = ans2 = ; for(int i = ; i < ; i++)
{
int ans = min(cnt1[i], cnt2[i]);
cnt1[i] -= ans;
cnt2[i] -= ans;
ans1 += ans;
} for(int i = ; i < ; i++)
{
int ans = min(cnt1[i], cnt2[i+'a'-'A']) + min(cnt2[i], cnt1[i+'a'-'A']);
ans2 += ans;
}
cout << ans1 << " " << ans2 << endl;
return ;
}
CodeForces 518B. Tanya and Postcard的更多相关文章
- 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) B. Tanya and Postcard 水题
B. Tanya and Postcard time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- CF Tanya and Postcard
Tanya and Postcard time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- CodeForces - 508D Tanya and Password(欧拉通路)
Description While dad was at work, a little girl Tanya decided to play with dad characters. She has ...
- codeforces 659C Tanya and Toys
题目链接:http://codeforces.com/problemset/problem/659/C 题意: n是已经有的数字,m是可用的最大数字和 要求选自己没有的数字,且这些数字的数字和不能超过 ...
- codeforces 659C . Tanya and Toys 二分
题目链接 将给出的已经有了的排序, 在前面加上0, 后面加上1e9+1. 然后对相邻的两项判断. 如果相邻两项之间的数的和小于m, 那么全都选上, m减去相应的值. 如果大于m, 那么二分判断最多能选 ...
- codeforces 508D . Tanya and Password 欧拉通路
题目链接 给你n个长度为3的子串, 这些子串是由一个长度为n+2的串分割得来的, 求原串, 如果给出的不合法, 输出-1. 一个欧拉通路的题, 将子串的前两个字符和后两个字符看成一个点, 比如acb, ...
- 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 ...
随机推荐
- wamp(win1064位家庭版+apache2.4.20+php5.5.37+mysql5.5.50)环境搭建
wamp环境搭建之软件准备 *php:http://windows.php.net/downloads/releases/php-5.5.37-Win32-VC11-x86.zip *apache:h ...
- VBA笔记(一)
开启VBA编程环境--VBE 方法一:按<Alt+F11>组合建 方法二:查看代码 宏设置 当然启用宏的设置方式不同,宏的启动方式也不一样. 首先打开"office 按钮&quo ...
- Android JSON、GSON、FastJson的封装与解析
声明: 1.本帖只提供代码,不深入讲解原理.如果读者想要深入了解,那就不要在这个帖子上浪费时间了 2.客户端用的是Google官方的Volley访问服务器,具体了解Volley请戳 这里 3.本帖三种 ...
- tyvj1189 盖房子
描述 永恒の灵魂最近得到了面积为n*m的一大块土地(高兴ING^_^),他想在这块土地上建造一所房子,这个房子必须是正方形的.但是,这块土地并非十全十美,上面有很多不平坦的地方(也可以叫瑕疵).这些瑕 ...
- idea之internal java compiler error
启动错误:Error:java: Compilation failed: internal java compiler error 解决:将圈选地方改为对应的jdk版本即可
- redis部署
下载软件 [root@localhost /]# wget http://download.redis.io/releases/redis-2.8.9.tar.gz 解压.编译.安装 [root@lo ...
- Uva 2319
理解:区域覆盖.假设该点在勘测半圆的边缘,求出与该点可在一个半圆的坐标范围l,r,然后,for 一次判断 #include<cstdio> #include<algorithm> ...
- python , angular js 学习记录【1】
1.日期格式化 Letter Date or Time Component Presentation Examples G Era designator Text AD y Year Year 199 ...
- python模块引用梳理
文件组织结构: t ├── __init__.py ├── main.py ├── t1 │ ├── A.py │ └── __init__.py └── t2 ├── B.py └── __ ...
- 数据库Sharding的基本思想和切分策略
一.基本思想 Sharding的基本思想就要把一个数据库切分成多个部分放到不同的数据库(server)上,从而缓解单一数据库的性能问题.不太严格的讲,对于海量数据的数据库,如果是因为表多而数据多,这时 ...