A13630 | Universal Solution
时间限制1s
内存限制256MB
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题目描述
Recently, you found a bot to play "Rock paper scissors" with. Unfortunately, the bot uses quite a simple algorithm to play: he has a string $s = s_1 s_2 \dots s_{n}$ of length $n$ where each letter is either R, S or P.
While initializing, the bot is choosing a starting index $pos$ ( $1 \le pos \le n$ ), and then it can play any number of rounds. In the first round, he chooses "Rock", "Scissors" or "Paper" based on the value of $s_{pos}$ :
- if $s_{pos}$ is equal to R the bot chooses "Rock";
- if $s_{pos}$ is equal to S the bot chooses "Scissors";
- if $s_{pos}$ is equal to P the bot chooses "Paper";
In the second round, the bot's choice is based on the value of $s_{pos + 1}$ . In the third round — on $s_{pos + 2}$ and so on. After $s_n$ the bot returns to $s_1$ and continues his game.
You plan to play $n$ rounds and you've already figured out the string $s$ but still don't know what is the starting index $pos$ . But since the bot's tactic is so boring, you've decided to find $n$ choices to each round to maximize the average number of wins.
In other words, let's suggest your choices are $c_1 c_2 \dots c_n$ and if the bot starts from index $pos$ then you'll win in $win(pos)$ rounds. Find $c_1 c_2 \dots c_n$ such that $\frac{win(1) + win(2) + \dots + win(n)}{n}$ is maximum possible.
While initializing, the bot is choosing a starting index $pos$ ( $1 \le pos \le n$ ), and then it can play any number of rounds. In the first round, he chooses "Rock", "Scissors" or "Paper" based on the value of $s_{pos}$ :
- if $s_{pos}$ is equal to R the bot chooses "Rock";
- if $s_{pos}$ is equal to S the bot chooses "Scissors";
- if $s_{pos}$ is equal to P the bot chooses "Paper";
In the second round, the bot's choice is based on the value of $s_{pos + 1}$ . In the third round — on $s_{pos + 2}$ and so on. After $s_n$ the bot returns to $s_1$ and continues his game.
You plan to play $n$ rounds and you've already figured out the string $s$ but still don't know what is the starting index $pos$ . But since the bot's tactic is so boring, you've decided to find $n$ choices to each round to maximize the average number of wins.
In other words, let's suggest your choices are $c_1 c_2 \dots c_n$ and if the bot starts from index $pos$ then you'll win in $win(pos)$ rounds. Find $c_1 c_2 \dots c_n$ such that $\frac{win(1) + win(2) + \dots + win(n)}{n}$ is maximum possible.
输入格式
The first line contains a single integer $t$ ( $1 \le t \le 1000$ ) — the number of test cases.
Next $t$ lines contain test cases — one per line. The first and only line of each test case contains string $s = s_1 s_2 \dots s_{n}$ ( $1 \le n \le 2 \cdot 10^5$ ; $s_i \in \{\text{R}, \text{S}, \text{P}\}$ ) — the string of the bot.
It's guaranteed that the total length of all strings in one test doesn't exceed $2 \cdot 10^5$ .
Next $t$ lines contain test cases — one per line. The first and only line of each test case contains string $s = s_1 s_2 \dots s_{n}$ ( $1 \le n \le 2 \cdot 10^5$ ; $s_i \in \{\text{R}, \text{S}, \text{P}\}$ ) — the string of the bot.
It's guaranteed that the total length of all strings in one test doesn't exceed $2 \cdot 10^5$ .
输出格式
For each test case, print $n$ choices $c_1 c_2 \dots c_n$ to maximize the average number of wins. Print them in the same manner as the string $s$ .
If there are multiple optimal answers, print any of them.
If there are multiple optimal answers, print any of them.
输入输出样例
输入 #1
3 RRRR RSP S
输出 #1
PPPP RSP R
In the first test case, the bot (wherever it starts) will always choose "Rock", so we can always choose "Paper". So, in any case, we will win all $n = 4$ rounds, so the average is also equal to $4$ .
In the second test case:
- if bot will start from $pos = 1$ , then $(s_1, c_1)$ is draw, $(s_2, c_2)$ is draw and $(s_3, c_3)$ is draw, so $win(1) = 0$ ;
- if bot will start from $pos = 2$ , then $(s_2, c_1)$ is win, $(s_3, c_2)$ is win and $(s_1, c_3)$ is win, so $win(2) = 3$ ;
- if bot will start from $pos = 3$ , then $(s_3, c_1)$ is lose, $(s_1, c_2)$ is lose and $(s_2, c_3)$ is lose, so $win(3) = 0$ ;
The average is equal to $\frac{0 + 3 + 0}{3} = 1$ and it can be proven that it's the maximum possible average. A picture from Wikipedia explaining "Rock paper scissors" game:

In the second test case:
- if bot will start from $pos = 1$ , then $(s_1, c_1)$ is draw, $(s_2, c_2)$ is draw and $(s_3, c_3)$ is draw, so $win(1) = 0$ ;
- if bot will start from $pos = 2$ , then $(s_2, c_1)$ is win, $(s_3, c_2)$ is win and $(s_1, c_3)$ is win, so $win(2) = 3$ ;
- if bot will start from $pos = 3$ , then $(s_3, c_1)$ is lose, $(s_1, c_2)$ is lose and $(s_2, c_3)$ is lose, so $win(3) = 0$ ;
The average is equal to $\frac{0 + 3 + 0}{3} = 1$ and it can be proven that it's the maximum possible average. A picture from Wikipedia explaining "Rock paper scissors" game:

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评测结果:Accepted