解法一 unique()函数将重复的元素放到vector的尾部 然后返回指向第一个重复元素的迭代器 再用erase函数擦除从这个元素到最后元素的所有的元素

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

vector<int> C;
int n, m;
int t;

int main()
{
    while (cin >> n >> m)
    {
        C.clear();
        for (int i = 1; i <= m + n; i ++ )
        {
            cin >> t;
            C.push_back(t);
        }
        sort(C.begin(), C.end());
        C.erase(unique(C.begin(), C.end()), C.end());
        for (int i = 0; i < C.size(); i ++ )
        {
            if (i != C.size() - 1) printf("%d ", C[i]);
            else printf("%d\n", C[i]);
        }
    }
    return 0;
}

解法二 利用set自带的去重功能

#include <iostream>
#include <set>

using namespace std;

int n, m;
set<int> st;

int main()
{
    while (cin >> n >> m)
    {
        st.clear();
        for (int i = 0; i < n + m; i ++ )
        {
            int x;
            cin >> x;
            st.insert(x);
        }
        int cnt = 0;
        for (set<int>::iterator it = st.begin(); it != st.end(); it ++ )
        {
            cout << *it;
            if (cnt != st.size() - 1) cout << ' ';
            cnt ++;
        }
        cout << endl;
    }
    return 0;
}