Binary Period
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
思路
這道題嘮叨半天，然後就被他套進去了，如果跟着題的意思走，搞來搞去發現情況越來越多，越來越難弄，而且dp好像也沒什麼用，其實如果用心讀題的話，s的長度小於等於2*t是一個暗示，暗示的很巧妙，但也是個坑，坑人坑的很慘。其實題目中的各種各樣的k的值根本不用管，1010的k值是2,10101010101010的k值還是2，如果一串字符爲0011，k爲4，而0101010101的k爲2，正好兩倍，一定能刪出目標來，一串字符，1的個數小於等於其長度，0的個數小於等於其長度，那麼2倍的長度，0,1，的個數分別爲長度，就行了。
哎，被坑的好慘！！！！！！！！
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
cin.tie(0);
ios::sync_with_stdio(0);
string s;
int t;
cin>>t;
while(t--)
{
cin>>s;
int len=s.length();
int i;
for( i=0;i<len-1;i++)
{
if(s[i]==s[i+1]) continue;
else break;
}
if(i==len-1)
{
for( i=1;i<=len;i++)
cout<<s[0];
cout<<endl;
}
else
{
for(int i=1;i<=len;i++)
{
cout<<"01";
}
cout<<endl;
}
}
return 0;
}