Children’s Game（String）

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最新推荐文章于 2025-07-17 19:18:54 发布

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本文介绍了一种针对儿童设计的数字游戏，通过特定算法找出由若干整数组合而成的最大可能数值。文章提供了一段C++代码实现，使用字符串比较及排序来解决该问题。

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Description

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 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 N 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

题意：

给出 N （1 ~ 50），代表有 N 个数，要求对这几个数合成一个数，输出字典序最大的。

思路：

本来想着数字按字典序由大到小输出就可以了，但是比如 9 和 90，90 字典序明显大于 9 若按这样输出的话应该是 909，但是事实上 990 可以构成更加大的。所以判断的时候应该判断 a + b > b + a （字符串连接号 ‘+’）。用 String，后 sort 即可。

AC：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string>

using namespace std;

bool cmp (string a, string b) { return a + b > b + a; }

int main () {
        int n;
        while (~scanf("%d", &n) && n) {
                string s[55];

                for (int i = 0; i < n; ++i)
                        cin >> s[i];

                sort(s, s + n, cmp);

                for (int i = 0; i < n; ++i)
                        cout << s[i];

                cout << endl;
        }
        return 0;
}