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;
     }
 }

 }