Elimination Round 2-A. Search for Pretty Integers

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本文介绍了一道编程题“寻找最小漂亮整数”的解决方法。题目要求从两个非零数字列表中找出至少包含每个列表一个元素的最小正整数。通过排序和遍历列表的方式，实现了有效的解决方案。

A. Search for Pretty Integers

题目链接：http://codeforces.com/contest/870/problem/A

Description

You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.

The second line contains n distinct digits a1, a2, ..., an (1 ≤ ai ≤ 9) — the elements of the first list.

The third line contains m distinct digits b1, b2, ..., bm (1 ≤ bi ≤ 9) — the elements of the second list.

Output
Print the smallest pretty integer.

Examples
input
2 3
4 2
5 7 6
output
25
input
8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
output
1
Note
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.

In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.
题意：给定两个数组，找到存在于两个数组的数组合成的最小数（重复算一个）。

题解；数据范围很小，直接暴力数组排序求解即可。

AC代码：

#include <iostream>
#include <algorithm>
using namespace std;
int n,m;
int min1,min2;
const int maxn = 15;
int a[maxn],b[maxn];
int main()
{
	cin>>n>>m;
	int min1=11;
	int min2=11;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	for(int j=1;j<=m;j++)
	{
		cin>>b[j];
	}
	sort(a+1,a+n+1);
	sort(b+1,b+m+1);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			if(a[i]==b[j]&&min1>a[i])
			{
				min2=i;
				min1=a[i];
			}
		}
	if(min2!=11)cout<<a[min2]<<endl;
	else
	{
		int t=0;
		if(a[1]<b[1])
		{
			t=a[1];
			a[1]=b[1];
			b[1]=t;
		}
		cout<<b[1]<<a[1]<<endl;

	}
	return 0;
}