codeforces Round #440 A Search for Pretty Integers【hash/排序】

A. Search for Pretty Integers

【题目链接】：http://codeforces.com/contest/872/problem/A

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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2个数组，求一个最小数，满足里面含有2个数组中至少一个数。注意特判俩个数组中含有相等的数的求看！

【分析】：方法一：分别hash一下每个数组的数，出现的标记为1，如果某个数在两个数组同时标记为1，说明在两个数组内都出现过，直接输出该数（如果两个数组出现同一个数字，纵使不是里面最小的，也要单独输出）。还要把两个数组内最小的mina和minb用递归找出来，再分别比较两者，最小放前，稍大置后。

方法二：排序。不用标记，直接升序排序。先判断两数组是否出现同一个数（二层循环枚举一下），否则分别将两数组第一个（最小）数记录，比较一下大小，小的在前大的在后。

【代码】：

#include <bits/stdc++.h>

using namespace std;
int main()
{
      int n,m,ma=10,mb=10;
      int x,y,a[10],b[10];//或者不用int，用bool
      memset(a,0,sizeof(a));
      memset(b,0,sizeof(b));//hash数组注意清0，否则结果错误
      cin>>n>>m;
      for(int i=1;i<=n;i++)
      {
          cin>>x;
          a[x]=1;
          ma=min(x,ma);
      }
      for(int i=1;i<=m;i++)
      {
          cin>>y;
          b[y]=1;
          mb=min(y,mb);
      }
      for (int i=1;i<=9;i++)
      {
        if (a[i] && b[i])
        {
            cout<<i<<endl;
            return 0;//真的好用，不能是break，否则还会顺序执行下去，会是不正确的输出
        }
      }
     cout<<min(ma,mb)<<max(ma,mb)<<endl;

  return 0;
}

 

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int n,m,a[10],b[10];
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;i++)
            cin>>a[i];
        sort(a+1,a+n+1);

        for(int i=1;i<=m;i++)
            cin>>b[i];
        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])
                {
                    cout<<a[i]<<endl;
                    return 0;
                }
            }
        }
        int ma=a[1];
        int mb=b[1];
        int minx=min(ma,mb);
        int maxx=max(ma,mb);
        cout<<minx<<maxx<<endl;
    }
}