B. Binary Period-白红宇

B. Binary Period

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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

111

110

outputCopy

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

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          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:; }}