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Codeforces Round 809(Div.2)

By xiaruize

A. Another String Minimization Problem

贪心，考虑优先改左侧的，如果左侧已经为A，则修改右侧的

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// Problem: A. Another String Minimization Problem
// Contest: Codeforces Round #809 (Div. 2)
// URL: https://codeforces.com/contest/1706/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, m;
int a[55];
string res;
int t;
// bool en;

void solve()
{
    res = "";
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
        res += "B";
    for (int i = 1; i <= n; i++)
    {
        if (a[i] < m - a[i] + 1)
        {
            if (res[a[i] - 1] == 'B')
                res[a[i] - 1] = 'A';
            else
                res[m - a[i]] = 'A';
        }
        else
        {
            if (res[m - a[i]] == 'B')
                res[m - a[i]] = 'A';
            else
                res[a[i] - 1] = 'A';
        }
    }
    cout << res << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

B. Making Towers

对于同样颜色的两块，如果两者之间隔偶数块(即坐标只差为奇数)，两块就可以满足上下堆叠，这需要统计奇偶性即可

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// Problem: B. Making Towers
// Contest: Codeforces Round #809 (Div. 2)
// URL: https://codeforces.com/contest/1706/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;

// bool st;
int t;
int n;
int c[N];
int cnt[N];
int b[N];
// bool en;

void solve()
{
    cin >> n;
    memset(cnt, 0, sizeof(cnt));
    memset(b, 0, sizeof(b));
    for (int i = 1; i <= n; i++)
    {
        cin >> c[i];
        if (!cnt[c[i]] || i % 2 != b[c[i]] % 2)
            cnt[c[i]]++, b[c[i]] = i;
    }
    for (int i = 1; i <= n; i++)
        cout << cnt[i] << ' ';
    cout << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

C. Qpwoeirut And The City

按nn的奇偶性分情况考虑

  1. n为奇数时 答案为把h2,h4,...,hn−1h_2,h_4,...,h_{n-1}调整到合适高度的最小代价

  2. n为偶数时 有且仅有一对相邻的两栋楼均不是cool的，且这一对一定出现在(2k−1,2k)(2k-1,2k)(k为正整数)

    枚举每一对可能的间隔，统计答案

    查询的过程可以通过前缀和，后缀和实现

    复杂度O(n)O(n)

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// Problem: C. Qpwoeirut And The City
// Contest: Codeforces Round #809 (Div. 2)
// URL: https://codeforces.com/contest/1706/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;

// bool st;
int t;
int n;
int a[N];
int pre[N];
int suf[N];
int r[N];
// bool en;

void solve()
{
    memset(pre, 0, sizeof(pre));
    memset(suf, 0, sizeof(suf));
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 2; i <= n; i++)
    {
        r[i] = max(max(a[i - 1], a[i + 1]) - a[i] + 1, 0ll);
        pre[i] = pre[i - 2] + r[i];
    }
    for (int i = n - 1; i >= 0; i--)
        suf[i] = suf[i + 2] + r[i];
    int res = INF;
    res = min(pre[n - 1], suf[2]);
    for (int i = 1; i <= n - 2; i += 2)
        res = min(res, pre[i - 1] + suf[i + 2]);
    if (n % 2 == 0)
        cout << res << endl;
    else
        cout << pre[n - 1] << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

D1. Chopping Carrots (Easy Version)

首先，可以考虑枚举最小值，此时要使得最大值最小，可以使用set维护每个数的商，此时可以lower_bound查询
时间复杂度: O(nk+na1logk)O(nk+na_1logk)

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// Problem: D1. Chopping Carrots (Easy Version)
// Contest: Codeforces Round #809 (Div. 2)
// URL: https://codeforces.com/contest/1706/problem/D1
// Memory Limit: 64 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, k;
int a[3005];
set<int> s[3005];
int minn = INF;
int t;
// bool en;

void solve()
{
    cin >> n >> k;
    minn = INF;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        s[i].clear();
        minn = min(minn, a[i]);
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= k; j++)
            s[i].insert(a[i] / j);
    }
    int res = INF;
    for (int i = 0; i <= minn; i++)
    {
        int mx = 0, mi = INF;
        for (int j = 1; j <= n; j++)
        {
            auto x = s[j].lower_bound(i);
            mx = max(mx, *x);
            mi = min(mi, *x);
        }
        res = min(res, mx - mi);
    }
    cout << res << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

D2. Chopping Carrots (Hard Version)

考虑优化上面的程序

处理m数组表示最小值为i时的最大值的最小值

时间复杂度： O(n)O(n)

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// Problem: D2. Chopping Carrots (Hard Version)
// Contest: Codeforces Round #809 (Div. 2)
// URL: https://codeforces.com/contest/1706/problem/D2
// Memory Limit: 64 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;

// bool st;
int n, k;
int a[N];
int s[N];
// bool en;

void solve()
{
    memset(s, 0, sizeof(s));
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        int tmp = INF;
        for (int j = 1; j <= min(a[i], k); j = (a[i] / (a[i] / j) + 1))
        {
            s[a[i] / j + 1] = max(s[a[i] / j + 1], tmp);
            tmp = a[i] / j;
        }
        s[0] = max(s[0], tmp);
    }
    int res = INF, now = 0;
    for (int i = 0; i <= a[1]; i++)
    {
        now = max(now, s[i]);
        res = min(res, now - i);
    }
    cout << res << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    int t;
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

未完待更