UVa 10905 - Children's Game（求多个正整数排列后，所得的新的数字的极值）

4thIIUCInter-University Programming Contest, 2005
A	Children’s Game
	Input: standard input Output: standard output
	Problemsetter: Md. Kamruzzaman

There are lots of number games for children. These games are pretty easy to play but not so easy to make. We will discuss about an interesting game here. Each player will be givenN positive integer. (S)He can make a big integer by appending those integers after one another. Such as if there are 4 integers as 123, 124, 56, 90 then the following integers can be made – 1231245690, 1241235690, 5612312490, 9012312456, 9056124123 etc. In fact 24 such integers can be made. But one thing is sure that 9056124123 is the largest possible integer which can be made.

You may think that it’s very easy to find out the answer but will it be easy for a child who has just got the idea of number?

Input

Each input starts with a positive integer N (≤ 50). In next lines there areN positive integers. Input is terminated by N = 0, which should not be processed.

Output

For each input set, you have to print the largest possible integer which can be made by appending all theN integers.

 

Sample Input	Output for Sample Input
4 123 124 56 90 5 123 124 56 90 9 5 9 9 9 9 9 0	9056124123 99056124123 99999

 

思路1：

贪心原则，为了让最终得到的数尽可能大，所以要让高位的数值尽可能大，直接按反字典序将数字串排序，并不正确，比如9和90排序后是90在前。

不过两个数字排列方式无非是一前一后，所以可以把两种排列的结果都比较下，这样做的的时间复杂度（只是两个数的比较）为O(3n)，同时要借助2n的空间（n为两数字的长度之和），但是写法很简单。

 

#include <iostream>

#include <cstdio>

#include <string>

#include <vector>

#include <algorithm>

using namespace std;



bool Cmp(const string &s, const string &t)

{

	string st = s + t;

	string ts = t + s;

	return st > ts;

}



int main(void)

{

	int n;

	while (cin >> n)

	{

		if (n == 0) break;



		vector<string> integers;

		int i;

		for (i = 0; i < n; ++i)

		{

			string str;

			cin >> str;

			integers.push_back(str);

		}



		sort(integers.begin(), integers.end(), Cmp);



		for (i = 0; i < n; ++i)

		{

			cout << integers[i];

		}

		cout << endl;

	}

	return 0;

}

思路2：

 

 

假设如下数字串：

A: a₁a₂a₃…a_n

B: b₁b₂b₃…b_m

AB: a₁a₂a₃…a_nb₁b₂b₃…b_m

BA: b₁b₂b₃…b_ma₁a₂a₃…a_n

判断两数字串谁应该放在前面时（这里把字符串应该放在前面的称为较大数字串），有两种情况

1、  A和B能够直接比较出大小

当比较到a_x和b_x时（x<=n且x<=m），分出大小。

2、  需要对AB和BA的值进行比较

AB: a₁a₂a₃…a_nb₁b₂b₃…b_m

BA: b₁b₂b₃…b_ma₁a₂a₃…a_n

比较到a_m和b_m时（m<n），仍然相等，这时说明

a₁a₂a₃…a_m= b₁b₂b₃…b_m①

接下来从AB式中的a_m+1和BA式中的a₁进行比较

AB₁:a_m+1a_m+2…a_nb₁b₂b₃…b_m

BA₁:a₁a₂a₃…a_n

用①式对BA₁进行替换，得到

AB₁:a_m+1a_m+2…a_nb₁b₂b₃…b_m

BA₁:b₁b₂b₃…b_m a_m+1a_m+2…a_n

可以发现AB₁和BA₁与AB和BA的形式完全相同，只是下标的值不同而已，所以这个比较过程是递归的。

因为整个比较过程中，AB和BA中总是有一个下标在向后走，所以时间复杂度为O(n)（n为两数字串的长度之和）。

 

#include <iostream>

#include <cstdio>

#include <string>

#include <vector>

#include <algorithm>

using namespace std;



bool Cmp(const string &s, const string &t)

{

	int i;

	for (i = 0; i < s.length() && i < t.length(); ++i)

	{

		if (s[i] != t[i]) return s[i] > t[i];

	}



	if (s.length() == t.length())

	{

		return true;

	}

	else if (i == s.length())

	{

		return Cmp(s, t.substr(i));

	}

	else

	{

		return Cmp(s.substr(i), t);

	}

}



int main(void)

{

	int n;

	while (cin >> n)

	{

		if (n == 0) break;



		vector<string> integers;

		int i;

		for (i = 0; i < n; ++i)

		{

			string str;

			cin >> str;

			integers.push_back(str);

		}



		sort(integers.begin(), integers.end(), Cmp);



		for (i = 0; i < n; ++i)

		{

			cout << integers[i];

		}

		cout << endl;

	}

	return 0;

}

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