Hiho----微软笔试题《Combination Lock》

Combination Lock

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Finally, you come to the interview room. You know that a Microsoft interviewer is in the room though the door is locked. There is a combination lock on the door. There are N rotators on the lock, each consists of 26 alphabetic characters, namely, 'A'-'Z'. You need to unlock the door to meet the interviewer inside. There is a note besides the lock, which shows the steps to unlock it.

Note: There are M steps totally; each step is one of the four kinds of operations shown below:

Type1: CMD 1 i j X: (i and j are integers, 1 <= i <= j <= N; X is a character, within 'A'-'Z')

This is a sequence operation: turn the ith to the jth rotators to character X (the left most rotator is defined as the 1st rotator)

For example: ABCDEFG => CMD 1 2 3 Z => AZZDEFG

Type2: CMD 2 i j K: (i, j, and K are all integers, 1 <= i <= j <= N)

This is a sequence operation: turn the ith to the jth rotators up K times ( if character A is turned up once, it is B; if Z is turned up once, it is A now. )

For example: ABCDEFG => CMD 2 2 3 1 => ACDDEFG

Type3: CMD 3 K: (K is an integer, 1 <= K <= N)

This is a concatenation operation: move the K leftmost rotators to the rightmost end.

For example: ABCDEFG => CMD 3 3 => DEFGABC

Type4: CMD 4 i j(i, j are integers, 1 <= i <= j <= N):

This is a recursive operation, which means:

If i > j:
	Do Nothing
Else:
	CMD 4 i+1 j
	CMD 2 i j 1

For example: ABCDEFG => CMD 4 2 3 => ACEDEFG

输入

1st line:  2 integers, N, M ( 1 <= N <= 50000, 1 <= M <= 50000 )

2nd line: a string of N characters, standing for the original status of the lock.

3rd ~ (3+M-1)th lines: each line contains a string, representing one step.

输出

One line of N characters, showing the final status of the lock.

提示

Come on! You need to do these operations as fast as possible.

 

样例输入

7 4
ABCDEFG
CMD 1 2 5 C
CMD 2 3 7 4
CMD 3 3
CMD 4 1 7

样例输出

HIMOFIN

EmacsNormalVim

 

 

import java.util.Scanner;


public class Main {
    
    
    public static void main(String[] argv){
        
        Scanner in = new Scanner(System.in);
        int M = in.nextInt();
        int N = in.nextInt();
        in.nextLine();
        String source = in.nextLine();
        String[] move= new String[N];
        for(int i=0; i<N;i++){
            move[i]=in.nextLine();
        }
        in.close();
        int[] int_Source = new int[M];    
        char[] c_Source = source.toCharArray(); 
        for(int i=0; i<M;i++){
            int_Source[i]= c_Source[i]-65;    
        }
        for(int i=0; i<N;i++){
            String[] temp = move[i].split(" ");
            switch(temp[1]){
            case "1":
                if(temp[4].length()==1)
                    oneMethod(int_Source,Integer.parseInt(temp[2]),Integer.parseInt(temp[3]),((int)(temp[4].toCharArray()[0]))-65);
                break;
            case "2":
                secondMethod(int_Source,Integer.parseInt(temp[2]),Integer.parseInt(temp[3]),Integer.parseInt(temp[4]));
                break;
            case "3":
                thirdMethod(int_Source,Integer.parseInt(temp[2]));
                break;
            case "4":
                fourthMethod(int_Source,Integer.parseInt(temp[2]),Integer.parseInt(temp[3]));
                break;    
            }
            /*
            for(int out:int_Source ){
                System.out.println(out+" ");
            }
            System.out.println();
            */
        }
        for(int out:int_Source ){
            System.out.print((char)(out+65));
        }
    }

    
    
    
    
public static void oneMethod(int[] s, int i, int j,int k){
    
    for(int p=i-1; p<j; p++){
        s[p]=k;
        s[p]=s[p]%26;
    }
    
}

public static void secondMethod(int[] s, int i, int j,int k){
    
    for(int p=i-1; p<j; p++){
        s[p]=s[p]+k;
        s[p]=s[p]%26;
    }    
}

public static void thirdMethod(int[] s, int i){
    int[] temp =new int[s.length];
    for(int q=0; q<s.length;q++){
        temp[q]=s[q];
    }
    for(int p=0; p<s.length; p++){
        if(i+p<s.length)
            s[p]=temp[i+p];
        else
            s[p]=temp[(i+p)%s.length];
        s[p]=s[p]%26;
    }
    
}

public static void fourthMethod(int[] s, int i, int j){
    
    for(int p=i-1;p<j;p++ ){        
        s[p]=s[p]+p-i+2;
        s[p]=s[p]%26;
    }
}

}