#include <bits/stdc++.h>
using namespace std;
int n, m, x;
signed main()
{
    while (~scanf("%d%d", &n, &m))
    {
        vector<int> a, b;
        for (int i = 0; i < n; i++)
        {
            cin >> x;
            a.push_back(x);
        }
        for (int i = 0; i < m; i++)
        {
            cin >> x;
            b.push_back(x);
        }
        a.insert(a.end(), b.begin(), b.end());
        sort(a.begin(), a.end());
        a.erase(unique(a.begin(), a.end()), a.end());
        for (int i = 0; i < a.size(); i++)
            cout << a[i] << " ";
        puts("");
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;

set<int > p;

int main() {
    int n, m,x;
    cin >> n >> m;
    while(n--) {
        cin >> x;
        p.insert(x);
    }
    while(m--) {
        cin >> x;
        p.insert(x);
    }
    for(auto it = p.begin(); it != p.end(); it++) {
        cout << *it << " ";
    }
    // system("pause");
    return 0;
}

此题为集合合并题，我们可以试着用Python解答，开一个set(),然后每次把a和b中的元素add到set()里，最后转换成列表输出即可 代码如下：

def getint():
    return list(map(int,input().split()))
while True:
    try:
        n,m=getint()
        a=getint()
        b=getint()
        s=set()
        for i in a:
            s.add(i)
        for i in b:
            s.add(i)
        s=list(s)
        s.sort()
        print(*s)
    except:
        break