uva 10905 - Children's Game

最新推荐文章于 2019-08-04 17:00:10 发布

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本文介绍了一种针对儿童设计的数字游戏，玩家需通过排列不同的整数来形成最大的可能数值。文章提供了完整的算法实现，包括输入输出示例，并讨论了通过自定义排序策略来简化问题的方法。

4^thIIUCInter-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

如果要自己找一个策略比较两个数哪个方前面还是比较容易出错的，不如直接写出这两个数的两种组合然后比较大小，利用sort重写比较函数即可，这样复杂度nlogn。

#include <stdio.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 50 + 5;
string a[maxn];
bool cmp(string x,string y){
    string tema = x + y;
    string temb = y + x;
    return tema > temb;
}

int main(){
    int n;
    while(scanf("%d",&n)){
        if(n == 0) break;
        for(int i = 0;i < n;i++) cin >> a[i];
        sort(a,a+n,cmp);
        for(int i = 0;i < n;i++) cout << a[i];
        printf("\n");
    }
    return 0;
}