Children’s Game - UVa 10905 拼接最大数

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本文介绍了一种针对儿童设计的数字游戏，通过给定的多个正整数，寻找由这些整数组合而成的最大可能数值。文章提供了一种排序算法思路，并附带了完整的AC代码实现。

4^th IIUC Inter-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 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

题意：给你n个数，问你用其拼接的最大的数是多少。

思路：用s1+s2>s2+s1的比较方式来排序，然后按序输出。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
string s[60];
bool cmp(string s1,string s2)
{
    return s1+s2>s2+s1;
}
int main()
{
    int n,i,j,k;
    while(~scanf("%d",&n) && n>0)
    {
        for(i=1;i<=n;i++)
           cin>>s[i];
        sort(s+1,s+1+n,cmp);
        for(i=1;i<=n;i++)
           cout<<s[i];
        printf("\n");
    }
}