ユニークビジョンプログラミングコンテスト2023 夏 (AtCoder Beginner Contest 312)
Ex
snukesnuke
AtCoder
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原题
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Problem Statement

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Takahashi is going to decide nicknames of $N$ people, person $1,\ldots,N$.

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Person $i$ wants a nickname $S_i$. To avoid giving the same nickname to multiple people, he is going to decide their nicknames as follows:

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  • For each $i=1,\ldots,N$ in order, decide person $i$'s nickname as follows:
  • Initialize a variable $k_i$ with $1$.
    • Repeatedly increment $k_i$ by one while the $k_i$-time repetition of $S_i$ is someone's nickname.
    • Let person $i$'s nickname be the $k_i$-time repetition of $S_i$.
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Find $k_1,\ldots$, and $k_N$ after deciding nicknames of the $N$ people.

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Constraints

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  • $N \geq 1$
  • $S_i$ is a string of length at least $1$ consisting of lowercase English letters.
  • The sum of lengths of $S_i$ is at most $2\times 10^5$.
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Input

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The input is given from Standard Input in the following format:

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N
S_1
\vdots
S_N

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Output

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Print $k_1,\ldots$, and $k_N$ resulting from deciding the nicknames of the $N$ people by the procedure in the problem statement.

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样例

Input #1
3
snuke
snuke
rng
Output #1
1 2 1
- First, he decides person $1$'s nickname. - Let $k_1=1$. - The $k_1$-time repetition of $S_1$ is `snuke`, which is nobody's nickname, so person $1$'s nickname is set to `snuke`. - Next, he decides person $2$'s nickname. - Let $k_2=1$. - The $k_2$-time repetition of $S_2$ is `snuke`, which is already a nickname of person $1$, so increment $k_2$ by one to make it $2$. - The $k_2$-time repetition of $S_2$ is `snukesnuke`, which is nobody's nickname, so person $2$'s nickname is set to `snukesnuke`. - Finally, he decides person $3$'s nickname. - Let $k_3=1$. - The $k_3$-time repetition of $S_3$ is `rng`, which is nobody's nickname, so person $3$'s nickname is set to `rng`. Thus, $k_1$, $k_2$, and $k_3$ result in $1$, $2$, and $1$, respectively.
Input #2
4
aa
a
a
aaa
Output #2
1 1 3 2
- Person $1$'s nickname is set to `aa`. - Person $2$'s nickname is set to `a`. - Person $3$'s nickname is set to `aaa`, because `a` and `aa` are already nicknames of someone else. - Person $4$'s nickname is set to `aaaaaa`, because `aaa` is already a nickname of someone else.
Input #3
5
x
x
x
x
x
Output #3
1 2 3 4 5
题目信息

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