Codeforces Global Round 16
B
MIN-MEX Cut
Codeforces
800
原题
← AC →
翻译中…

A binary string is a string that consists of characters $0$ and $1$.

翻译中…

Let $\operatorname{MEX}$ of a binary string be the smallest digit among $0$, $1$, or $2$ that does not occur in the string. For example, $\operatorname{MEX}$ of $001011$ is $2$, because $0$ and $1$ occur in the string at least once, $\operatorname{MEX}$ of $1111$ is $0$, because $0$ and $2$ do not occur in the string and $0 \lt 2$.

翻译中…

A binary string $s$ is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.

翻译中…

A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

翻译中…

What is the minimal sum of $\operatorname{MEX}$ of all substrings pieces can be?

翻译中…

Input

翻译中…

The input consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases. Description of the test cases follows.

翻译中…

Each test case contains a single binary string $s$ ($1 \le |s| \le 10^5$).

翻译中…

It's guaranteed that the sum of lengths of $s$ over all test cases does not exceed $10^5$.

翻译中…

Output

翻译中…

For each test case print a single integer — the minimal sum of $\operatorname{MEX}$ of all substrings that it is possible to get by cutting $s$ optimally.

翻译中…

Note

翻译中…

In the first test case the minimal sum is $\operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1$.

翻译中…

In the second test case the minimal sum is $\operatorname{MEX}(1111) = 0$.

翻译中…

In the third test case the minimal sum is $\operatorname{MEX}(01100) = 2$.

翻译中…

样例

Input #1
6
01
1111
01100
101
0000
01010
Output #1
1
0
2
1
1
2
InputCopy OutputCopy
题目信息
题号
1566B
时间限制
1 s
内存限制
256 MB
标签
位运算
构造
动态规划
贪心

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。