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

#include <iostream>
#include <stdlib.h>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <queue>

using namespace std;

void cmd1(string &str, int i, int j, char c) {
    for (int m = i; m <= j; ++m) {
        str[m - 1] = c;
    }
}

void cmd2(string &str, int i, int j, int k) {
    for (int m = i; m <= j; ++m) {
        int v = (str[m - 1] - 'A' + k) % 26;
        str[m - 1] = 'A' + v;
    }
}

void reverse(string &str, int i, int j) {
    while (i < j) {
        swap(str[i], str[j]);
        i++, j--;
    }
}

void cmd3(string &str, int k) {
    reverse(str, 0, k - 1);
    reverse(str, k, str.length() - 1);
    reverse(str, 0, str.length() - 1);
}

void cmd4(string &str, int i, int j) {
    if (i > j) return;
    //cmd4(str, i + 1, j);
    //cmd2(str, i, j, 1);
    for (int m = i; m <= j; ++m) {
        int v = (str[m - 1] - 'A' + m - i + 1) % 26;
        str[m - 1] = 'A' + v;
    }
}

int main(int argc, char** argv) {
    int n, m;
    cin >> n >> m;
    string input;
    cin >> input;

    for (int step = 0; step < m; ++step) {
        string cmd; 
        int type, i, j, k; 
        char c;
        cin >> cmd >> type;
        if (type == 1) {
            cin >> i >> j >> c;
            cmd1(input, i, j, c);
        } else if (type == 2) {
            cin >> i >> j >> k;
            cmd2(input, i, j, k);
        } else if (type == 3) {
            cin >> k;
            cmd3(input, k);
        } else if (type == 4) {
            cin >> i >> j;
            cmd4(input, i, j);
        }
    //cout << input << endl;
    }

    cout << input << endl;
    return 0;
}

老是TLE。估计是递归开销太大了。分析了一下改成迭代的了。也不知道对不对。。。