4th IIUC Inter-University Programming Contest, 2005 / UVa 10905 Children's Game (等长转化技巧)

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本文介绍了一种针对儿童设计的数字游戏，玩家需利用给定的整数组合出最大的可能数值。文章详细解释了游戏规则，并提供了一段C++代码实现，通过调整输入数字的顺序来确保最终组合出的最大整数。

10905 - Children's Game

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=512&page=show_problem&problem=1846

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 given N 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 are N 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 the Nintegers.

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

怎么比较90和9的大小？——把两个数的长度变成一样即可。（详见代码）

完整代码：

/*0.045s*/

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100;

struct Numstr
{
	int len;
	char s[N];
	bool operator < (const Numstr &n) const
	{
		return strcmp(s, n.s) > 0; ///从大到小
	}
} p[50];

int main(void)
{
	int n, l, x, maxl;
	while (scanf("%d", &n), n)
	{
		maxl = 0;
		for (int i = 0; i < n; ++i)
		{
			scanf("%s", p[i].s);
			l = strlen(p[i].s);
			p[i].len = l;
			if (l > maxl)
				maxl = l;
		}
		++maxl;///为什么加了这个就能A????????
		for (int i = 0; i < n; ++i)
		{
			x = l = p[i].len;
			while (x < maxl)
			{
				p[i].s[x] = p[i].s[x - l];///循环复制
				x++;
			}
			p[i].s[maxl] = '\0';
		}
		sort(p, p + n);
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < p[i].len; ++j)
				putchar(p[i].s[j]);
		putchar('\n');
	}
}