luogu#P3491. [POI 2009] SLW-Words
[POI 2009] SLW-Words
题目描述
Let be a function acting on strings composed of the digits 0 and 1. The function transforms the string by replacing (independently and concurrently) every digit 0 with 1 and every digit 1 with the string "10". For example , (i.e. assigns an empty string to the empty string). Note that is an injection, or a one - to - one function. By we denote the function composed with itself times. In particular, is the identity function .
We are interested in the strings of the form for . This sequence begins with the following strings:
"0", "1", "10", "101", "10110", "10110101".
We call the string a substring of the string if it occurs in as a contiguous (i.e. one - block) subsequence. A sequence of integers is given. Your task is to check whether a string of the form $h^{k_1}("0") \cdot h^{k_2}("0") \cdots h^{k_n}("0")$ is a substring of for some .
输入格式
The first line of the standard input contains a single integer , , denoting the number of test units. The first line of each test unit's description contains one integer , . The second line of each description holds non - negative integers , separated by single spaces. The sum of the numbers in the second line of any test unit description does not exceed 10000000.
输出格式
Your programme should print out lines to the standard output, one for each test unit. Each line corresponding to a test unit should contain one word: TAK (yes in Polish - if $h^{k_1}("0") \cdot h^{k_2}("0") \cdots h^{k_n}("0")$ is a substring of for some in that test unit, or NIE (no in Polish) otherwise.
题目大意
题目描述
函数 作用于由数字 和 组成的字符串,其通过将每一个数字 替换为 ,将每一个数字 替换为字符串 来独立且同时地变换字符串 。
例如:
其中例二即把空字符串赋值给空字符串。
注意, 是一个一对一的函数。
我们用 表示函数 使用 次。
特别的, 。
我们对 的 形式的字符串感兴趣,这个序列为:
$$``0", ``1", ``10", ``101", ``10110", ``10110101", \ldots $$如果字符串 在 中以一个连续的(即一个块)子串出现,我们就称 为字符串 的子串。给出整数 的序列。你的任务是检查是否有 ,使 $h^{k_1}(``0") \cdot h^{k_2}(``0") \cdot \ldots \cdot h^{k_n}(``0") $ 形式的字符串是 的子串。
输入内容
输入的第一行包含一个整数 表示样例数, 。每个样例输入两行,一行 , ,另一行为 个非负整数 ,用单个空格分隔。
输出内容
您的程序应该输出 行,每个测试输出一行。如果存在 使 $h^{k_1}(``0") \cdot h^{k_2}(``0") \cdot \ldots \cdot h^{k_n}(``0") $ 是 的子串,就输出 TAK
(波兰语中的 yes ),如果不存在 ,则输出 NIE
(波兰语中的 no )。
Translated By in_young_ren
2
2
1 2
2
2 0
TAK
NIE