微软在线笔试题2015

描述

We say a stringis beautiful if it has the equal amount of 3 or more continuous letters (inincreasing order.)

Here are someexample of valid beautiful strings: "abc", "cde","aabbcc", "aaabbbccc".

Here are someexample of invalid beautiful strings: "abd", "cba","aabbc", "zab".

Given a stringof alphabets containing only lowercase alphabets (a-z), output "YES"if the string contains a beautiful sub-string, otherwise output "NO".

输入

The first linecontains an integer number between 1 and 10, indicating how many test cases arefollowed.

For each testcase: First line is the number of letters in the string; Second line is thestring. String length is less than 10MB.

输出

For each testcase, output a single line "YES"/"NO" to tell if the stringcontains a beautiful sub-string.

提示

Huge input. SlowIO method such as Scanner in Java may get TLE.

样例输入

4

3

abc

4

aaab

6

abccde

3

abb

样例输出

YES

NO

YES

NO

时间限制:8000ms

单点时限:1000ms

内存限制:256MB

描述

You are given atxt file, which is performance logs of a single-threaded program.

Each line hasthree columns as follow:

[Function Name][TimeStamp] [Action]

[FunctionName]is a string of length between 1~255

[TimeStamp]format is hh:mm:ss

Valid values for"Action" column are START or END, marking the start or end of afunction call.

Each functionwill only be called once.

Output thedepth-first traversal result of the call graph with the total time of eachfunction call. However, sometimes the performance log isn‘t correct and at thattime you just need to output "Incorrect performance
log".

输入

The input onlycontains 1 case, first line is a positive number N representing the number oflogs(1 <= N <= 20000), then there are N lines in next, each line is thelog info containing [Function Name] [TimeStamp] [Action],
[Function Name] is astring, you can assume the [Function Name] is distinct and the length between1~255.

输出

Output thedepth-first traversal result of the call graph with the total time of eachfunction call for the correct performance, or output "Incorrectperformance log".

提示

A call graph isa directed graph that represents calling relationships between subroutines in acomputer program.

Call graph forthe sample input is shown as below:

Another sampletest case.

样例输入

8

FuncA 00:00:01 START

FuncB 00:00:02 START

FuncC 00:00:03 START

FuncC 00:00:04 END

FuncB 00:00:05 END

FuncD 00:00:06 START

FuncD 00:00:07 END

FuncA 00:00:08 END

样例输出

FuncA 00:00:07

FuncB 00:00:03

FuncC 00:00:01

FuncD 00:00:01

窗体顶端

EmacsNormalVim

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

We define thematching contents in the strings of strA and strB as common substrings of thetwo strings. There are two additional restrictions on the common substrings.

The firstrestriction here is that every common substring‘s length should not be less than3.  For example:

strA: abcdefghijklmn

strB: ababceghjklmn

The matchingcontents in strA and strB are substrings ("abc", "jklmn").Note that though "e" and "gh" are common substrings of strAand strB, they are not matching content because their lengths are less than 3.

The secondrestriction is that the start indexes of all common substrings should bemonotone increasing. For example:

strA: aaabbbbccc

strB: aaacccbbbb

The matchingcontents in strA and strB are substrings ("aaa", "bbbb").Note that though "ccc" is common substring of strA and strB and haslength greater than 3, the start indexes of ("aaa", "bbbb","ccc") in strB are
(0, 6, 3), which is not monotone increasing.

输入

Two lines. Thefirst line is strA and the second line is strB. Both strA and strB are oflength less than 2100.

输出

The length ofmatching contents (the sum of the lengths of the common substrings).

样例输入

abcdefghijklmn

ababceghjklmn

样例输出

8

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Finally, youcome to the interview room. You know that a Microsoft interviewer is in theroom though the door is locked. There is a combination lock on the door. Thereare 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 anote besides the lock, which shows the steps to unlock it.

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

Type1: CMD1 i j X: (i and j are integers, 1 <= i <= j <= N; X is a character,within ‘A‘-‘Z‘)

This is asequence operation: turn the ith to the jth rotators to character X (the leftmost rotator is defined as the 1st rotator)

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

Type2: CMD2 i j K: (i, j, and K are all integers, 1 <= i <= j <= N)

This is asequence operation: turn the ith to the jth rotators up K times ( if characterA 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: CMD3 K: (K is an integer, 1 <= K <= N)

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

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

Type4: CMD4 i j(i, j are integers, 1 <= i <= j <= N):

This is arecursive operation, which means:

If i > j:

DoNothing

Else:

CMD4 i+1 j

CMD2 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: astring of N characters, standing for the original status of the lock.

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

输出

One line of Ncharacters, showing the final status of the lock.

提示

Come on! Youneed 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

窗体底端