微软2014实习生及校招秋令营技术类职位，在线编程题目及解答。

题目2 : K-th string

时间限制:10000ms

单点时限:1000ms

内存限制:256MB



Description

Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If such a string doesn’t exist, or the input is not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.
Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.
Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output.
Output

For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”.


样例输入

3
2 2 2
2 2 7
4 7 47

样例输出

0101
Impossible
01010111011

题目解答：

#include <iostream>
#include <String>

using namespace std;

void reorder(string input);

void main()
{
	string input;
	while(cin >> input)
	{
		reorder(input);
	}
}

void reorder(string input)
{
	int *ptiShowTimes = new int[36]();
	int finishFlag = true;

	//统计每一个字母出现的次数
	for (int i = 0; i < input.length(); i++)
	{
		if ((input[i] < '0') || (input[i] > 'z') || ((input[i] > '9') && (input[i] < 'a')))
		{
			cout << "<invalid input string>" << endl;
			return;
		}
		if ((input[i] >= '0') && (input[i] <= '9'))
		{
			ptiShowTimes[input[i]-'0']++;
		}
		else
		{
			ptiShowTimes[input[i]-'a'+10]++;
		}

	}

	do
	{
		finishFlag = true;
		for (int i = 0; i < 36; i++)
		{
			if (0 != ptiShowTimes[i])
			{
				finishFlag = false;
				if (i < 10)
				{
					cout << (char)(i + '0');
					ptiShowTimes[i]--;
				}
				else
				{
					cout << (char)('a' + i - 10);
					ptiShowTimes[i]--;
				}
			}
		}
	}while (!finishFlag);

	cout << endl;
}

题目3 : Reduce inversion count

时间限制:10000ms

单点时限:1000ms

内存限制:256MB



Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count.

Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

Example:
Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2
InversionCountOfSwap({3, 1, 2})=>
{
 InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1
 InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1
 InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1
}
Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.
Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.


样例输入

3,1,2
1,2,3,4,5

样例输出

1
0

题目解答：

#include <sstream>
#include <vector>
#include <map>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

using namespace std;
int calInversionCount(vector<int>::iterator begin, int *ptrSmallerNum, int *ptrBiggerNum, int dataLen);
int getMostInversionCount(int *ptrSmallerNum, int dataLen);

void main()
{
	int dataLen = 0;
	string str = "";
	while (getline(cin, str))
	{
		vector<int> data;
		stringstream ss(str);
		string tmp;
		while (getline(ss, tmp, ','))
			data.push_back(atoi(tmp.c_str()));

		dataLen = data.size();

		//申请两个数组，用于存放小于它的个数和大于它的个数
		int *ptrBiggerNum = new int[dataLen]();
		int *ptrSmallerNum = new int[dataLen]();

		//计算出每一个元素小于它的个数和大于它的个数
		int inversionOldCount = calInversionCount(data.begin(), ptrSmallerNum, ptrBiggerNum, dataLen);

		//获得inversionCount个数最大的元素
		int iMostInverCountIndex = getMostInversionCount(ptrSmallerNum, dataLen);

		//获得比其他元素小最多次数的元素
		int iSmallestInverCountIndex = getMostInversionCount(ptrBiggerNum, dataLen);

		//将上述两个元素进行交换
		swap(data[iMostInverCountIndex], data[iSmallestInverCountIndex]);

		//计算新的inversioncount
		int inversionNewCount = calInversionCount(data.begin(), ptrSmallerNum, ptrBiggerNum, dataLen);

		if (inversionNewCount < inversionOldCount)
		{
			cout << inversionNewCount << endl;
		}
		else
		{
			cout << inversionOldCount << endl;
		}

		delete[] ptrBiggerNum;
		delete[] ptrSmallerNum;
	}

}

int getMostInversionCount(int *ptrSmallerNum, int dataLen)
{
	int iMostInversionCount = ptrSmallerNum[0];
	int i = 0;
	int result = 0;
	for (; i < dataLen; i++ )
	{
		if (ptrSmallerNum[i] >= iMostInversionCount)
		{
			iMostInversionCount = ptrSmallerNum[i];
			result = i;
		}
	}
	return result;
}

int calInversionCount(vector<int>::iterator begin, int *ptrSmallerNum, int *ptrBiggerNum, int dataLen)
{
	int inversionCount = 0;

	int i = 0;
	int j = 0;

	for (; i < dataLen; i++)
	{
		for (j = i + 1; j < dataLen; j++)
		{
			if (*(begin+j) < *(begin+i))
			{
				//统计出每一个成员的小于它的数和大于它的数
				ptrSmallerNum[i]++;
				ptrBiggerNum[j]++;
				inversionCount++;
			}
		}
	}
	return inversionCount;
}

题目4 : Most Frequent Logs

时间限制:10000ms

单点时限:3000ms

内存限制:256MB



Description

In a running system, there are many logs produced within a short period of time, we‘d like to know the count of the most frequent logs.

Logs are produced by a few non-empty format strings, the number of logs is N(1<=N<=20000), the maximum length of each log is 256.

Here we consider a log same with another when their edit distance (see note) is <= 5.

Also we have a) logs are all the same with each other produced by a certain format string b) format strings have edit distance  5 of each other.

Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length.

Note edit distance is the minimum number of operations (insertdeletereplace a character) required to transform one string into the other, please refer to

http://en.wikipedia.org/wiki/Edit_distance for more details.
Input

Multiple lines of non-empty strings.
Output

The count of the most frequent logs.



样例输入

Logging started for id:1
Module ABC has completed its job
Module XYZ has completed its job
Logging started for id:10
Module ? has completed its job


样例输出

3

题目解答：

做不动了，后面有解答的时候再补上吧！