题目1 : String reorder
时间限制:10000ms 单点时限:1000ms 内存限制:256MB
Description
For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’).
Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met,
1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order).
2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment.
Your program should output string “<invalid input string>” when the input contains any invalid characters (i.e., outside the ‘0’-‘9′ and ‘a’-‘z’ range).
Input
Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters.
Output
For each case, print exactly one line with the reordered string based on the criteria above.
- 样例输入
aabbccdd 007799aabbccddeeff113355zz 1234.89898 abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee
- 样例输出
abcdabcd 013579abcdefz013579abcdefz <invalid input string> abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa
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！
