Codeforces 1099 C. Postcard-字符串处理(Codeforces Round #530 (Div. 2))
1 second
256 megabytes
standard input
standard output
Andrey received a postcard from Irina. It contained only the words "Hello, Andrey!", and a strange string consisting of lowercase Latin letters, snowflakes and candy canes. Andrey thought that this string is an encrypted message, and decided to decrypt it.
Andrey noticed that snowflakes and candy canes always stand after the letters, so he supposed that the message was encrypted as follows. Candy cane means that the letter before it can be removed, or can be left. A snowflake means that the letter before it can be removed, left, or repeated several times.
For example, consider the following string:

This string can encode the message «happynewyear». For this, candy canes and snowflakes should be used as follows:
- candy cane 1: remove the letter w,
- snowflake 1: repeat the letter p twice,
- candy cane 2: leave the letter n,
- snowflake 2: remove the letter w,
- snowflake 3: leave the letter e.

Please note that the same string can encode different messages. For example, the string above can encode «hayewyar», «happpppynewwwwwyear», and other messages.
Andrey knows that messages from Irina usually have a length of kk letters. Help him to find out if a given string can encode a message of kkletters, and if so, give an example of such a message.
The first line contains the string received in the postcard. The string consists only of lowercase Latin letters, as well as the characters «*» and «?», meaning snowflake and candy cone, respectively. These characters can only appear immediately after the letter. The length of the string does not exceed 200200.
The second line contains an integer number kk (1≤k≤2001≤k≤200), the required message length.
Print any message of length kk that the given string can encode, or «Impossible» if such a message does not exist.
hw?ap*yn?eww*ye*ar
12
happynewyear
ab?a
2
aa
ab?a
3
aba
ababb
5
ababb
ab?a
1
Impossible
题意就是给你一个字符串,其中?可以有两种操作,1.将前面的字母放到当前位置,其实就是去掉?2.去掉前面的字母,其实就是去掉前面的字母去掉?
*的有三种操作,1.将前面的字母放到当前位置,就是去掉* 2.去掉前面的字母,就是去掉前面的字母和* 3.前面的字母可以重复任意次
假设是abc*,可以是abc,可以是ab,可以是abcccccc,就是这样的,给你一个字符串,问你能不能将字符串变成要求的长度,最后的字符串不能有?和*。
直接将纯字母的长度先找出来,然后判断再进行操作就可以了。
代码:
//C
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn=1e5+; char s[]; int main()
{
scanf("%s",s);
int n;
cin>>n;
int len=strlen(s);
int candy=,snow=;
for(int i=;i<len;i++){
if(s[i]=='?') candy++;
else if(s[i]=='*') snow++;
}
int l=len-candy-snow;
if(l==n){
for(int i=;i<len;i++){
if(s[i]!='?'&&s[i]!='*') cout<<s[i];
}
cout<<endl;
}
else if(l<n&&snow>){
int c;
for(int i=;i<len;i++){
if(s[i]=='?') s[i]='#';
if(s[i]=='*'){
if(l<n){
c=n-l;
s[i]='!';
//cout<<s[i]<<endl;
l=n;
}
else{
s[i]='#';
}
}
}
for(int i=;i<len;i++){
if(s[i]!='#'){
if(s[i]=='!'){
for(int j=;j<c;j++)
cout<<s[i-];
}
else cout<<s[i];
}
}
cout<<endl;
}
else if(l>n&&l-(candy+snow)<=n){
for(int i=;i<len;i++){
if((s[i]=='?'||s[i]=='*')&&l>n){
s[i-]='#';
s[i]='#';
l--;
}
else if(s[i]=='?'||s[i]=='*')
s[i]='#';
}
for(int i=;i<len;i++){
if(s[i]!='#') cout<<s[i];
}
cout<<endl;
}
else cout<<"Impossible"<<endl;
}
Codeforces 1099 C. Postcard-字符串处理(Codeforces Round #530 (Div. 2))的更多相关文章
- Codeforces 1099 D. Sum in the tree-构造最小点权和有根树 贪心+DFS(Codeforces Round #530 (Div. 2))
D. Sum in the tree time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces 1099 B. Squares and Segments-思维(Codeforces Round #530 (Div. 2))
B. Squares and Segments time limit per test 1 second memory limit per test 256 megabytes input stand ...
- Codeforces 1099 A. Snowball-暴力(Codeforces Round #530 (Div. 2))
A. Snowball time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...
- Codeforces Round #530 (Div. 2) A,B,C,D
A. Snowball 链接:http://codeforces.com/contest/1099/problem/A 思路:模拟 代码: #include<bits/stdc++.h> ...
- Codeforces Round #530 (Div. 2) F (树形dp+线段树)
F. Cookies 链接:http://codeforces.com/contest/1099/problem/F 题意: 给你一棵树,树上有n个节点,每个节点上有ai块饼干,在这个节点上的每块饼干 ...
- Codeforces Round #530 (Div. 2):D. Sum in the tree (题解)
D. Sum in the tree 题目链接:https://codeforces.com/contest/1099/problem/D 题意: 给出一棵树,以及每个点的si,这里的si代表从i号结 ...
- Codeforces Round #530 (Div. 2) F 线段树 + 树形dp(自下往上)
https://codeforces.com/contest/1099/problem/F 题意 一颗n个节点的树上,每个点都有\(x[i]\)个饼干,然后在i节点上吃一个饼干的时间是\(t[i]\) ...
- Codeforces Round #530 (Div. 2)F Cookies (树形dp+线段树)
题:https://codeforces.com/contest/1099/problem/F 题意:给定一个树,每个节点有俩个信息x和t,分别表示这个节点上的饼干个数和先手吃掉这个节点上一个饼干的的 ...
- Codeforces Round #530 (Div. 2) Solution
A. Snowball 签. #include <bits/stdc++.h> using namespace std; ], d[]; int main() { while (scanf ...
随机推荐
- PIP 批量更新改为清华这边的镜像更新
之前pip批量更新的时候发现有些包无法更新,而且速度也特别慢,今天尝试了下清华的镜像,速度是真快 # coding=utf-8import pipfrom subprocess import call ...
- mysql5.6以上(适用5.7)免安装版本 终极配置
1.解压你的mysql5.6 我解压的位置是D:\Program Files\mysql--winx64,你可以随意放在任何位置,不建议解压到C盘 2.来到你解压的文件根目录下,新建一个my.ini文 ...
- 【poj2947】高斯消元求解同模方程组【没有AC,存代码】
题意: p start enda1,a2......ap (1<=ai<=n)第一行表示从星期start 到星期end 一共生产了p 件装饰物(工作的天数为end-start+1+7*x, ...
- 回顾一下MyBatis逆向工程——自动生成代码
前言 最近做的项目(SSM+Shiro)的数据库表已经创建完成,一共有15张表,如果我们一个个去写pojo/bean的代码以及各种sql语句的话未免太过麻烦而且很容易出错,这个时候我们就需要MyBat ...
- 深入分析Spring 与 Spring MVC容器(山东数漫江湖)
1 Spring MVC WEB配置 Spring Framework本身没有Web功能,Spring MVC使用WebApplicationContext类扩展ApplicationContext, ...
- HDU 1070 Milk (模拟)
题目链接 Problem Description Ignatius drinks milk everyday, now he is in the supermarket and he wants to ...
- 网络协议之HTTP协议
HTTP协议详解(真的很经典) 转自:http://blog.csdn.net/gueter/archive/2007/03/08/1524447.aspx Author :Jeffrey 引言 HT ...
- SD卡 模拟SPI总线控制流程
SD卡为移动设备提供了安全的,大容量存储解决方法.它本身可以通过两种总线模式和MCU进行数据传输,一种是称为SD BUS的4位串行数据模式,另一种就是大家熟知的4线SPI Bus模式.一些廉价,低端的 ...
- 64_g4
gnatcoll-2014-10.fc26.x86_64.rpm 28-Feb-2017 17:44 1738266 gnatcoll-devel-2014-10.fc26.i686.rpm 28-F ...
- 10 个打造 React.js App 的最佳 UI 框架
10 个打造 React.js App 的最佳 UI 框架 在本文中,我们将分享一些助你打造 React.js App 最佳的 UI 框架.它们具备你所需要的基本 React 组件,以及易用的 API ...