代码源挑战赛R12

R12A

#include <bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n - i; j++)
            cout << " ";
        for (int j = 1; j <= i * 2 - 1; j++)
            cout << "#";
        cout << "\n";
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n - 1; j++)
            cout << " ";
        cout << "#\n";
    }

    return 0;
}

R12B

辗转相减

时间复杂度 $O(n)$，$100$ 分

#include <bits/stdc++.h>
using namespace std;
int a, b, x, y;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> a >> b;
    x = y = 0;
    while (a != b)
    {
        if (a > b)
            x++, a -= b;
        else
            y++, b -= a;
    }
    cout << x << " " << y << "\n";
    return 0;
}

辗转相除

时间复杂度 $O(\log n)$，$100$ 分

#include <bits/stdc++.h>
using namespace std;
int a, b, x, y;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> a >> b;
    x = y = 0;
    while (a != b && a && b)
    {
        if (a > b)
        {
            x += a / b;
            a %= b;
        }
        else
        {
            y += b / a;
            b %= a;
        }
    }
    if (a == 0)
        x--;
    if (b == 0)
        y--;
    cout << x << " " << y << "\n";
    return 0;
}

R12C

链表

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
const int MAXM = 300000;
int n, m;
int pre[MAXN + 5];
int nxt[MAXN + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        pre[i] = i - 1;
        nxt[i] = i + 1;
    }
    pre[1] = 0;
    nxt[n] = 0;
    int head = 1;
    for (int i = 1; i <= m; i++)
    {
        int xi;
        cin >> xi;
        if (xi == head)
            continue;
        // xi 从链表中删除
        nxt[pre[xi]] = nxt[xi];
        pre[nxt[xi]] = pre[xi];
        // xi 放到链表头
        pre[head] = xi;
        nxt[xi] = head;
        pre[xi] = 0;
        head = xi;
    }
    for (int i = head; i != 0; i = nxt[i])
        cout << i << " ";
    return 0;
}

标记权值重新排序

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
const int MAXM = 300000;
int n, m;
pair<int, int> num[MAXN + 5];
int x[MAXM + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        num[i] = {i, i};
    int now = 1;
    for (int i = 1; i <= m; i++)
    {
        cin >> x[i];
        num[x[i]].first = --now;
    }
    sort(num + 1, num + n + 1);
    for (int i = 1; i <= n; i++)
        cout << num[i].second << " ";
    return 0;
}

标记每个数还在不在

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
const int MAXM = 300000;
int n, m;
int a[MAXN + MAXM + 5];
int pos[MAXN + MAXM + 5];
bool ok[MAXN + MAXM + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    int now = 1 + m;
    for (int i = 1; i <= n; i++)
    {
        a[i + m] = i;
        ok[i + m] = true;
        pos[i] = i + m;
    }
    for (int i = 1; i <= m; i++)
    {
        int xi;
        cin >> xi;
        ok[pos[xi]] = false;
        a[--now] = xi;
        ok[now] = true;
        pos[xi] = now;
    }
    for (int i = now; i <= n + m; i++)
        if (ok[i])
            cout << a[i] << " ";
    return 0;
}

离线处理

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
const int MAXM = 300000;
int n, m;
int x[MAXM + 5];
bool flag[MAXN + 5]; // 每个数有没有输出
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
        cin >> x[i];
    for (int i = m; i >= 1; i--)
    {
        if (!flag[x[i]])
        {
            cout << x[i] << " ";
            flag[x[i]] = true;
        }
    }
    for (int i = 1; i <= n; i++)
        if (!flag[i])
            cout << i << " ";
    return 0;
}

R12D

枚举每种答案组合

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MOD = 998244353;
const int MAXN = 3'000'000;
int n;
vector<int> d;
int cnt[MAXN + 5];
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int now = gcd(i, n);
        cnt[now]++;
        if (cnt[now] == 1)
            d.push_back(now);
    }
    int ans = 0;
    for (int i = 0; i < d.size(); i++)
    {
        int x = cnt[d[i]];
        for (int j = 0; j < d.size(); j++)
        {
            int y = cnt[d[j]];
            ans += x * y * gcd(d[i], d[j]);
            ans %= MOD;
        }
    }
    cout << ans;
    return 0;
}