Codeforces Round #440 A. Search for Pretty Integers

最新推荐文章于 2022-11-22 23:15:14 发布

原创最新推荐文章于 2022-11-22 23:15:14 发布 · 447 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#CodeForce #ACM

ACM简单题专栏收录该内容

50 篇文章

订阅专栏

本文介绍了一种算法，用于寻找包含两个非零数字列表中各至少一个元素的最小正整数。通过输入两个列表的长度及具体内容，算法能快速找出满足条件的最小整数。

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.

水题，找最小的数要么在两个数组里头出现过，要么由两个数组中的数组成

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int a[15],b[15];
long long ff(long long q1,long long q2)
{
    int s=0;
    long long t=q2;
    while(t)
    {
        t=t/10;
        s++;
    }
    return q1*((long long)(pow(10,s)))+q2;
}
int main() {
    int x,y;
    cin>>x>>y;
    for (int i=1; i<=x; i++) {
        cin>>a[i];
    }
    for (int i=1; i<=y; i++) {
        cin>>b[i];
    }
    sort(a+1, a+x+1);
    sort(b+1, b+y+1);
    bool f=false;
    long long t=1e9+7;
    for(int i=1;i<=x;i++)
    {
        for (int j=1; j<=y; j++) {
            if(a[i]==b[j])
            {
                f=true;
                t=a[i];
                break;
            }
        }
        if(f)
            break;
    }
    t=min(t,min(ff(a[1],b[1]),ff(b[1],a[1])));
    cout<<t<<endl;
    // insert code here...
    //std::cout << "Hello, World!\n";
    return 0;
}