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Codeforces Round 927 (Div. 3)

时间：2024-02-19 19:45:12浏览次数：24  
标签：int ll Codeforces long back vector using Div 927

Codeforces Round 927 (Div. 3)

A - Thorns and Coins

解题思路:

出现连续两个障碍之前，所有金币都能拿到。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    s = ' ' + s;
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == '@')
        {
            ans++;
        }
        else if (s[i] == '*')
        {
            if (i - 1 > 0 && s[i - 1] == s[i])
            {
                break;
            }
        }
    }
    cout << ans << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

B - Chaya Calendar

解题思路:

遍历，找到大于已过当前年的最小倍数。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    vector<ll> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    ll cur = 1;
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
        {
            cur = a[i] + 1;
        }
        else
        {
            ll k = (cur + (a[i] - 1)) / a[i];
            cur = k * a[i] + 1;
        }
    }
    cout << cur - 1 << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C - LR-remainders

解题思路:

从全部开始删不好搞，我们先按步骤走到最后一点，然后反着加起来。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    ll n, m;
    cin >> n >> m;
    vector<ll> a(n + 1, 1);
    ll ans = 1;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    string s;
    cin >> s;
    int l = 1;
    int r = n;
    int idx = 0;
    for (auto c : s)
    {
        if (c == 'L')
        {
            l++;
        }
        else
        {
            r--;
        }
    }
    reverse(s.begin(), s.end());
    vector<int> v;
    for (auto c : s)
    {
        if (c == 'L')
        {
            l--;
            ans = ans * a[l] % m;
        }
        else
        {
            r++;
            ans = ans * a[r] % m;
        }
        v.push_back(ans);
    }
    for (int i = v.size() - 1; i >= 0; i--)
    {
        cout << v[i] << ' ';
    }
    cout << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D - Card Game

解题思路:

模拟吐了。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    string jok;
    cin >> jok;
    map<char, int> q;
    q['C'] = 1;
    q['D'] = 2;
    q['H'] = 3;
    q['S'] = 4;
    map<int, char> inv;
    inv[1] = 'C';
    inv[2] = 'D';
    inv[3] = 'H';
    inv[4] = 'S';
    vector<vector<int>> cards(10, vector<int>());
    for (int i = 1; i <= 2 * n; i++)
    {
        string t;
        cin >> t;
        int x = q[t[1]];
        cards[x].push_back(t[0] - '0');
    }
    for (int i = 1; i <= 4; i++)
    {
        sort(cards[i].begin(), cards[i].end());
    }
    // for (int i = 1; i <= 4; i++)
    // {
    //     for (auto x : cards[i])
    //     {
    //         cout << x << ' ';
    //     }
    //     cout << endl;
    // }
    vector<pair<string, string>> ans;
    for (int i = 1; i <= 4; i++)
    {
        // cerr << i << endl;
        if (q[jok[0]] == i)
        {
        }
        else
        {
            int len = cards[i].size();
            for (int j = len - 1; j >= 1; j -= 2)
            {
                // cerr << j << ' ';
                int a = cards[i].back();
                cards[i].pop_back();
                int b = cards[i].back();
                cards[i].pop_back();
                char cb = char('0' + b);
                string tb;
                tb.push_back(cb);
                tb.push_back(inv[i]);
                char ca = '0' + a;
                string ta;
                ta.push_back(ca);
                ta.push_back(inv[i]);
                ans.push_back({tb, ta});
            }
            // cerr << endl;
        }
    }
    for (int i = 1; i <= 4; i++)
    {
        if (q[jok[0]] != i)
        {
            while (!cards[i].empty())
            {
                int a = cards[i].back();
                cards[i].pop_back();
                if (cards[q[jok[0]]].size())
                {
                    int b = cards[q[jok[0]]].back();
                    cards[q[jok[0]]].pop_back();
                    char cb = char('0' + b);
                    string tb;
                    tb.push_back(cb);
                    tb.push_back(jok[0]);
                    char ca = '0' + a;
                    string ta;
                    ta.push_back(ca);
                    ta.push_back(inv[i]);
                    ans.push_back({ta, tb});
                }
                else
                {
                    puts("IMPOSSIBLE");
                    return;
                }
            }
        }
    }
    if (cards[q[jok[0]]].size())
    {
        int len = cards[q[jok[0]]].size();
        for (int j = len - 1; j >= 1; j -= 2)
        {
            int a = cards[q[jok[0]]].back();
            cards[q[jok[0]]].pop_back();
            int b = cards[q[jok[0]]].back();
            cards[q[jok[0]]].pop_back();
            char cb = char('0' + b);
            string tb;
            tb.push_back(cb);
            tb.push_back(inv[q[jok[0]]]);
            char ca = '0' + a;
            string ta;
            ta.push_back(ca);
            ta.push_back(inv[q[jok[0]]]);
            ans.push_back({tb, ta});
        }
    }
    for (int i = 1; i <= 4; i++)
    {
        if (cards[i].size())
        {
            puts("IMPOSSIBLE");
            return;
        }
    }
    for (auto t : ans)
    {
        cout << t.fi << ' ' << t.se << endl;
    }
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E - Final Countdown

解题思路:

举例:\(12345 \to 12345 + 1234 + 123 + 12 + 1 = 13715\)

显然，全部加起来，边除边模往上进位即可。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    for (int i = 0; i < n; i++)
    {
        if (s[i] != '0')
        {
            s = s.substr(i);
            break;
        }
    }
    n = s.size();
    s = ' ' + s;
    vector<ll> pre(n + 10, 0);
    for (int i = 1; i <= n; i++)
    {
        pre[i] = pre[i - 1] + (s[i] - '0');
    }
    string ans;
    ll res = 0;
    for (int i = n; i > 0; i--)
    {
        char c = (pre[i] % 10) + '0';
        ans += c;
        if (i > 1)
        {
            pre[i - 1] += pre[i] / 10;
        }
        else
        {
            res += pre[i] / 10;
        }
    }
    while (res)
    {
        char c = (res % 10) + '0';
        res /= 10;
        ans += c;
    }
    reverse(ans.begin(), ans.end());
    cout << ans << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

F - Feed Cats

解题思路:

\(dp[i]:考虑前i个点的选取，能得到的最大线段数。\)

  • 如果选第\(i\)个点，那么我们要找到覆盖点\(i\)的线段中，最小的左端点位置\(idx\)，\(dp[i] \leftarrow dp[idx - 1] + cnt\)。其中，\(cnt\)为经过点\(i\)的线段数量。
  • 如果不选第\(i\)个点，就是考虑前\(i\)个点得到的最大答案\(dp[i] \leftarrow dp[i - 1]\)。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<vector<int>> l(n + 1, vector<int>());
    vector<int> d(n + 10);
    for (int i = 1; i <= m; i++)
    {
        int a, b;
        cin >> a >> b;
        l[b].push_back(a);
        d[a]++;
        d[b + 1]--;
    }
    for (int i = 1; i <= n; i++)
    {
        d[i] += d[i - 1];
    }
    // 左侧最远可影响i处的点。
    vector<int> w(n + 10, n + 1);
    for (int i = n; i; i--)
    {
        w[i] = min(i, w[i + 1]);
        for (auto x : l[i])
        {
            w[i] = min(w[i], x);
        }
    }
    vector<int> dp(n + 1);
    for (int i = 1; i <= n; i++)
    {
        dp[i] = max(dp[i - 1], dp[w[i] - 1] + d[i]);
    }
    cout << dp[n] << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

G - Moving Platforms

解题思路:

\((u,v)\)之间连通情况。

根据题意，\(l_u + ks_u \equiv l_v + ks_v \mod h\)

\[k(s_u - s_v) \equiv l_v - l_u \mod h \\ \]

\[k(s_u - s_v) + h y = l_v - l_u \]

扩展欧几里得解不定方程，得到\((k_0, y_0)\)。\(gcd((s_u - s_v),h) = d\)

根据裴蜀定理,判断是否有解，如果无解，就无法连边。

特解\((k_0,y_0)\)，通解为\((k_0 + t\frac{h}{d},y_0 - t\frac{s_u - s_v}{d})\)

求最小非负整数\(k\)，以及合法时间点间隔\(\frac{h}{d}\)。

然后跑最短路。转移时找到最近合法可转移的时间进行转移。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const ll inf = 1ll << 50;
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll g = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
}

struct node
{
    ll v, st, step;
};

void solve()
{
    ll n, m, h;
    cin >> n >> m >> h;
    vector<ll> l(n + 1), s(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> l[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i];
    }
    vector<vector<node>> e(n + 1, vector<node>());
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        ll c = l[v] - l[u];
        ll a = s[u] - s[v];
        ll b = h;
        ll x, y;
        ll d = exgcd(a, b, x, y);
        if (c % d)
        {
            continue;
        }
        ll step = abs(b / d);
        ll k = c / d % step;
        x = x * k % step;
        x = ((x % step) + step) % step;
        e[u].push_back((node){v, x, step});
        e[v].push_back((node){u, x, step});
    }
    priority_queue<pii, vector<pii>, greater<pii>> q;
    vector<ll> dist(n + 1, inf);
    dist[1] = 0;
    vector<bool> vis(n + 1);
    q.push({dist[1], 1});
    while (q.size())
    {
        auto t = q.top();
        q.pop();
        int u = t.se;
        if (vis[u])
        {
            continue;
        }
        vis[u] = true;
        for (auto x : e[u])
        {
            ll v = x.v;
            ll sp = x.st;
            ll step = x.step;
            // 找到最小合法转移时间
            if (dist[u] > sp)
            {
                sp += (dist[u] - sp + (step - 1)) / step * step;
            }
            if (dist[v] > sp + 1)
            {
                dist[v] = sp + 1;
                q.push({dist[v], v});
            }
        }
    }
    if (dist[n] >= inf)
    {
        puts("-1");
    }
    else
    {
        cout << dist[n] << endl;
    }
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

标签：int,ll,Codeforces,long,back,vector,using,Div,927
From： https://www.cnblogs.com/value0/p/18021816

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