Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination) A. Search for Pretty Integers

A. Search for Pretty Integers

Problem Statement

    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

Example 1
    Input
        2 3
        4 2
        5 7 6
    Output
        25
Example 2
    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

    给出两个序列，序列中的数∈[1,9]，要求找出一个最小的数使得这个数的数位上的数在两个序列中都出现过。

思路

    我们很容易想到，如果两个list中有相同的数的话，那只要输出最小的同时出现在两个list里的数就行了，因为1位数肯定比两位数小（这不是废话么）。如果没有的话，那么就分别拿出两个list中的最小数，让小的放在前面，大的放在后面，就可以组成最小的符合要求的数了。

Code

#pragma GCC optimize(3)#include<bits/stdc++.h>using namespace std;typedef long long ll;inline void readInt(int &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}inline void readLong(ll &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}/*================Header Template==============*/int a[2][10];int n,m;vector<int> s;int main() {    cin>>n>>m;    for(int i=1;i<=n;i++)        cin>>a[0][i];    for(int i=1;i<=m;i++)        cin>>a[1][i];    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)            if(a[0][i]==a[1][j])                s.push_back(a[0][i]);            else                s.push_back(min(a[0][i],a[1][j])*10+max(a[0][i],a[1][j]));    sort(s.begin(),s.end());    cout<<s[0]<<endl;    return 0;}