牛客Round35

A

////直接截取
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
int n, m;
void solve()
{
    string s;
    cin >> s;
    int len = s.size() / 2;
    cout << s.substr(0, len) << '\n';
    cout << s.substr(len, len);
}

int main()
{
    solve();
}

B

//排序后直接两两比较
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
void solve()
{
    int n;
    cin >> n;
    vector<int> a(2 * n);
    for (int &x : a)
        cin >> x;
    sort(a.begin(), a.end());
    for (int i = 0; i < 2 * n; i += 2)
    {
        if (a[i] != a[i + 1])
        {
            cout << -1;
            return;
        }
    }
    for (int i = 0; i < 2 * n; i += 2)
    {
        cout << a[i] << " ";
    }
    cout << endl;
    for (int i = 1; i < 2 * n; i += 2)
    {
        cout << a[i] << " ";
    }
}

int main()
{
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--)
        solve();
}

//map做法
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
constexpr int N = 2e6 + 10;

int n, m;
int a[N], b[N], c[N];
void solve()
{
    cin >> n;
    map<int, int> mp;
    int x;
    rep(i, 1, 2 * n)
    {
        cin >> x;
        ++mp[x];
    }
    int cnt = 0;
    for (auto [u, v] : mp)
    {
        if (v & 1)
        {
            cout << -1 <<'\n';
            return;
        }
        int len = v / 2;
        rep(i, 1, len)
        {
            a[++cnt] = u;
            b[cnt] = u;
        }
    }
    rep(i, 1, n)
    {
        if (i > 1)
            cout << ' ';
        cout << a[i];
    }
    cout << '\n';
    rep(i, 1, n)
    {
        if (i > 1)
            cout << ' ';
        cout << a[i];
    }
    cout << '\n';
}

int main()
{
    cout.tie(0);
    solve();
}

C

// 排序+双指针，从左边开始找
//记录最大值
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
void solve()
{
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    sort(a.begin(), a.end());
    int l = 0;
    int dis = 0;
    for (int i = 0; i < n; i++)
    {
        while (a[i] - a[l] > k)
            l++;
        dis = max(dis, i - l + 1);
    }
    cout << 1.0 * dis / n;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--)
        solve();
}

D

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
void solve()
{
    int n;
    cin >> n;
    vector<pair<int, int>> a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].first;
        a[i].second = i + 1;
    }
    sort(a.begin(), a.end());
    vector<int> need(n + 1, 0);
    vector<pair<int, int>> vp;
    vector<int> vis(n, 0);
    for (int i = 0; i < n; i++)
    {
        if (a[i].first > n)
            continue;
        if (need[a[i].first] == 0)
        {
            need[a[i].first] = 1;
            vis[i] = 1;
            continue;
        }
    }
    int p = 1;
    for (int i = 0; i < n; ++i)
    {
        if (vis[i] == 0)
        {
            while (need[p] == 1)
                p++;
            need[p] = 1;
            vp.push_back({a[i].second, p});
        }
    }
    cout << vp.size() << '\n';
    for (auto p : vp)
    {
        cout << p.first << " " << p.second << endl;
    }
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--)
        solve();
}
{
    solve();
}

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
constexpr int N = 2e6 + 10, md = 1e9 + 7;
typedef pair<int,int> PII;
int n, m;
int a[N];

bool vis[N]; 
void solve()
{
    cin >> n;
    rep(i, 1, n)
    {
        cin >> a[i];
    }
    set<int> st;
    rep(i, 1, n)
    {
        st.insert(i);
    }
    vector<int> pos;
    rep(i, 1, n)
    {
        if (a[i] <= n)
        {
            if (st.find(a[i]) != st.end())
            {
                st.erase(a[i]);
            }
            else
            {
                pos.push_back(i);
            }
        }
        else
        {
            pos.push_back(i);
        }
    }
    vector<PII> ans;
    int cnt = 0;
    for (int v : st)
    {
        ans.push_back({pos[cnt++], v});
    }
    cout << ans.size() << '\n';
    for (auto [u, v] : ans)
    {
        cout << u << ' ' << v << '\n';
    }
}

int main()
{

    solve();
}

E

//分层，可以先把每一层的节点连接到上一层的第一个节点
//再看看然后第i层还需不需要连接
//差不多就是一个凑的思路
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
void solve()
{
    int n, m;
    cin >> n >> m;
    //输入
    vector<vector<int>> adj(n);
    //存每一层的
    vector<int> a(n);
    //存一开始的
    for (int &x : a)
        cin >> x;
        //输入a[i],0-n-1
    for (int i = 1; i < n; i++)
        if (a[i] == 0)
        {
            cout << -1;
            return;
            //其他边到第一条边不可能是0
        }
    for (int i = 0; i < n; i++)
        adj[a[i]].push_back(i);
        //放进去每一层的点
    while (adj.back().size() == 0)
        adj.pop_back();
        //如果末尾是空的，就不断Pop

    vector<pair<int, int>> ans;
    //存储答案
    for (int i = 1; i < adj.size(); i++)
    {
        if (adj[i].size() == 0)
        {
            cout << -1;
            //如果出现了所谓的断层
            //那么久弄掉
            return;
        }
    }

    for (int i = 1; i < adj.size(); i++)
    {
        for (int x : adj[i])
        {
            ans.push_back({adj[i - 1][0], x});
        }
    }
    //先放每一层连接上一层的点
    //连了n-1次
    m -= n - 1;
    if (m < 0)
    {
        cout << -1;
        return;
    }
    //不够？那么结束
    for (int i = 0; i < adj.size() - 1 && m; i++)
    {
        for (int j = 0; j < adj[i].size() && m; j++)
        {
            for (int k = 0; k < adj[i + 1].size() && m; k++)
            {
                if (j == 0)
                    continue;
                ans.push_back({adj[i][j], adj[i + 1][k]});
                m--;
            }
        }
    }
    //开始放每一层连接上一层的其他，注意m随时推出
    for (int i = 0; i < adj.size() && m; i++)
    {
        for (int j = 0; j < adj[i].size() && m; j++)
        {
            for (int k = j + 1; k < adj[i].size() && m; k++)
            {
                ans.push_back({adj[i][j], adj[i][k]});
                m--;
            }
        }
    }
    //开始连接本层，注意m随时推出
    if (m)
    {
        cout << -1;
        return;
    }
    //如果已经都连完了，还能连，就输了
    for (auto p : ans)
    {
        cout << p.first + 1 << " " << p.second + 1 << endl;
    }
    //输出答案，注意+1
}

int main()
{
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--)
        solve();
}