博客
关于我
B. Binary Period
阅读量:538 次
发布时间:2019-03-09

本文共 2257 字,大约阅读时间需要 7 分钟。

Let’s say string s has period k if si=si+k for all i from 1 to |s|−k (|s| means length of string s) and k is the minimum positive integer with this property.

Some examples of a period: for s=“0101” the period is k=2, for s=“0000” the period is k=1, for s=“010” the period is k=2, for s=“0011” the period is k=4.

You are given string t consisting only of 0’s and 1’s and you need to find such string s that:

String s consists only of 0’s and 1’s;

The length of s doesn’t exceed 2⋅|t|;
String t is a subsequence of string s;
String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t=“011” is a subsequence of s=“10101”.

Input

The first line contains single integer T (1≤T≤100) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains string t (1≤|t|≤100) consisting only of 0’s and 1’s.

Output

Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.

Example

inputCopy
4
00
01
111
110
outputCopy
00
01
11111
1010
Note
In the first and second test cases, s=t since it’s already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.

In the third test case, there are shorter optimal solutions, but it’s okay since we don’t need to minimize the string s. String s has period equal to 1.

题意

给你一个二进制字符串t,求一个字符串s,满足t是s的子序列,s的长不超过t的两倍,且s的周期最小
思路
因为是二进制字符串,我们让它01或10交替(t的每一个字符变为两个),就可以保证T=2,且s的长度不超过2倍t的长度。
当然,如果字符串t中只有‘1’或者‘0’的话,T=1,我们直接让s=t即可

#include
#include
#include
#include
#include
#include
//#include
using namespace std;typedef long long LL;int main(){ int n; cin >> n; while (n--) { string t; cin >> t; if (t.size() <= 2) //t的大小≤2,显然直接输出即可 { cout << t << endl; } else { int cnt0 = 0,cnt1=0; for (int i = 0; i < t.size(); i++) { if (t[i] == '0') cnt0++; else cnt1++; } if (cnt0 == t.size() || cnt1 == t.size()) { cout << t << endl; goto ca; } char h = t[0], hh=1-(t[0]-'0')+'0'; for (int i = 1; i <= t.size(); i++) { cout << h << hh; } cout << endl; } ca:; }}

转载地址:http://xboiz.baihongyu.com/

你可能感兴趣的文章
MYSQL CONCAT函数
查看>>
multiprocessing.Pool:map_async 和 imap 有什么区别?
查看>>
MySQL Connector/Net 句柄泄露
查看>>
multiprocessor(中)
查看>>
mysql CPU使用率过高的一次处理经历
查看>>
Multisim中555定时器使用技巧
查看>>
MySQL CRUD 数据表基础操作实战
查看>>
multisim变压器反馈式_穿过隔离栅供电:认识隔离式直流/ 直流偏置电源
查看>>
mysql csv import meets charset
查看>>
multivariate_normal TypeError: ufunc ‘add‘ output (typecode ‘O‘) could not be coerced to provided……
查看>>
MySQL DBA 数据库优化策略
查看>>
multi_index_container
查看>>
MySQL DBA 进阶知识详解
查看>>
Mura CMS processAsyncObject SQL注入漏洞复现(CVE-2024-32640)
查看>>
Mysql DBA 高级运维学习之路-DQL语句之select知识讲解
查看>>
mysql deadlock found when trying to get lock暴力解决
查看>>
MuseTalk如何生成高质量视频(使用技巧)
查看>>
mutiplemap 总结
查看>>
MySQL DELETE 表别名问题
查看>>
MySQL Error Handling in Stored Procedures---转载
查看>>