Codeforces Round #327 (Div. 2)B(逻辑)
2 seconds
256 megabytes
standard input
standard output
The name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding — an active marketing strategy, that includes a set of measures to change either the brand (both for the company and the goods it produces) or its components: the name, the logo, the slogan. They decided to start with the name.
For this purpose the corporation has consecutively hired m designers. Once a company hires the i-th designer, he immediately contributes to the creation of a new corporation name as follows: he takes the newest version of the name and replaces all the letters xi by yi, and all the letters yi by xi. This results in the new version. It is possible that some of these letters do no occur in the string. It may also happen that xi coincides with yi. The version of the name received after the work of the last designer becomes the new name of the corporation.
Manager Arkady has recently got a job in this company, but is already soaked in the spirit of teamwork and is very worried about the success of the rebranding. Naturally, he can't wait to find out what is the new name the Corporation will receive.
Satisfy Arkady's curiosity and tell him the final version of the name.
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the length of the initial name and the number of designers hired, respectively.
The second line consists of n lowercase English letters and represents the original name of the corporation.
Next m lines contain the descriptions of the designers' actions: the i-th of them contains two space-separated lowercase English letters xi and yi.
Print the new name of the corporation.
6 1
police
p m
molice
11 6
abacabadaba
a b
b c
a d
e g
f a
b b
cdcbcdcfcdc
In the second sample the name of the corporation consecutively changes as follows:






题意:
给出一个长为n的字符串a[],进行m次操作;
每次操作描述为将字符a改成字符b,将字符b改成字符a;
思路:
开一个字符数组str存储a...z26个字符。。直接对str里面的字符进行交换操作。。
然后用a[i]-"a"做下标。。str[a[i]-"a"]即为交换后的字符。。
代码:
#include <bits/stdc++.h>
using namespace std; int main(void)
{
std::ios::sync_with_stdio(false);
int n, m;
string a;
cin >> n >> m >> a;
char str[];
for(int i=; i<; i++)
str[i]=i+'a';
while(m--)
{
char a, b;
cin >> a >> b;
if(a==b) continue;
for(int i=; i<; i++)
{
if(str[i]==a) str[i]=b;
else if(str[i]==b) str[i]=a;
}
}
for(int i=; i<a.size(); i++)
cout << str[a[i]-'a'];
cout << endl;
return ;
}
Codeforces Round #327 (Div. 2)B(逻辑)的更多相关文章
- Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题
A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...
- Codeforces Round #327 (Div. 2) E. Three States BFS
E. Three States Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/probl ...
- Codeforces Round #327 (Div. 2) D. Chip 'n Dale Rescue Rangers 二分 物理
D. Chip 'n Dale Rescue Rangers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/co ...
- Codeforces Round #327 (Div. 2) C. Median Smoothing 找规律
C. Median Smoothing Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/p ...
- Codeforces Round #327 (Div. 2) B. Rebranding 水题
B. Rebranding Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem ...
- Codeforces Round #327 (Div. 1), problem: (A) Median Smoothing
http://codeforces.com/problemset/problem/590/A: 在CF时没做出来,当时直接模拟,然后就超时喽. 题意是给你一个0 1串然后首位和末位固定不变,从第二项开 ...
- Codeforces Round #327 (Div. 2) B. Rebranding C. Median Smoothing
B. Rebranding The name of one small but proud corporation consists of n lowercase English letters. T ...
- Codeforces Round #327 (Div. 2)
题目传送门 水 A - Wizards' Duel 题目都没看清就写了,1e-4精度WA了一次... /************************************************ ...
- Codeforces Round #327 (Div. 2)-Wizards' Duel
题意: 在一条长度为l的走廊,两个人站在走廊的左右两端分别以p,q的速度走来,问他们相遇时离左端的距离是多少? 思路: 非常简单的暴力题,不解释. 代码如下: #include <iostrea ...
随机推荐
- javascript高级程序设计---document节点
document节点是文档的根节点,每张网页都有自己的document节点,window.document就是指向这个节点.只要浏览器开始载入文档,这个节点就开始了 对于HTML文档来说,docume ...
- Javascript高级程序设计——基本包装类型
既然js中的基本类型没有属性和方法那么为什么对字符串进行subString()方法可以呢?基本类型不应该没有方法的吗? 这就是基本包装类型啦! ECMAScript提供了三个特殊的引用类型,Boole ...
- OpenGL基础图形的绘制
例一:绘制五角星 设五角星5个顶点分别为ABCDE,顶点到坐标轴原点距离为r,ABCDE各点用r表示,分别为 A(0,r); B(-r*sin(2/5*Pi),r*cos(2/5*Pi)); C(-r ...
- Opencv二值图像的分布直方图
Mat img; ]; int main() { VideoCapture video(); if (!video.isOpened()) { ; } Mat img; Mat img1, img2, ...
- 9 DelayQueueEntry 延时队列节点类——Live555源码阅读(一)基本组件类
这是Live555源码阅读的第一部分,包括了时间类,延时队列类,处理程序描述类,哈希表类这四个大类. 本文由乌合之众 lym瞎编,欢迎转载 http://www.cnblogs.com/oloroso ...
- Python文件操作题
1.如何用Python删除一个文件 os.remove(filename)或者os.unlink(filename). 2.Python如何copy一个文件 shutil模块里有一个copyfile函 ...
- 算法题解之math类题
Bulb Switcher 灯泡开关 思路:除了平方数以外,其他所有位置的灯泡最终都被开关了偶数次,因此最终都为0.问题等价于求1~n中平方数的个数. public class Solution { ...
- c# winform UI + python底层的一点尝试
鉴于python做winform之类的UI比较弱.于是想结合C#的winform 和 python的底层开发(windows平台). 尝试做了一个RSS阅读器.在这里:http://download. ...
- SDL播放声音
extern "C" { #include <libavcodec/avcodec.h> #include <libavformat/avformat.h> ...
- C# 复制(深拷贝、浅拷贝)
Object.MemberwiseClone 方法 创建当前 Object 的浅表副本. protected Object MemberwiseClone() MemberwiseClone 方法创建 ...