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Good Bye 2013---B. New Year Present

來源：程序員人生發布時間：2014-11-09 08:14:20 閱讀次數：2435次

The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.

Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.

Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.

Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.

Input

The first line contains integer n (2?≤?n?≤?300) ― the number of wallets. The next line contains n integers a1,?a2,?...,?an (0?≤?ai?≤?300).

It is guaranteed that at least one ai is positive.

Output

Print the sequence that consists of k (1?≤?k?≤?106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.

As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.

Sample test(s)

input
2
1 2

output
PRPLRP

input
4
0 2 0 2

output
RPRRPLLPLRRRP

解題思路：對第個位置，先判斷糖果數是不是大于0，若大于，則先輸出'P',再重復判斷，若等于0，則輸出'R',同時往后移。當1個位置上糖果數大于1時，輸出第2個及以上糖果時，有兩種方式回到原地，'PL'和‘LP’，這個要判斷1下在邊界的情況，要保證機器人不能出界。

AC代碼：

#include <iostream> #include <cstdio> using namespace std; int a[305]; int main(){ // freopen("in.txt","r",stdin); int n; while(cin>>n){ for(int i=0; i<n; i++) cin>>a[i]; for(int j=0; j<n; j++){ int k = 0; while(a[j] > 0){ if(k){ if(j == n⑴) cout<<"LR"; else cout<<"RL"; } cout<<"P"; a[j] --; k ++; } if(j < n⑴) cout<<"R"; } cout<<endl; } return 0; }

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Good Bye 2013---B. New Year Present