牛客小白月赛99

A

取黑色和彩色的最小即可。

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve()
{
    ll a, b, x, y;
    cin >> a >> b >> x >> y;
    cout << a * min(x, y) + b * y << '\n';
}
int main()
{
    int Case;
    cin >> Case;
    while (Case--)
        solve();
    return 0;
}

B

贪心，尽量对半砍。

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#include <bits/stdc++.h>
#define int long long
using namespace std;
using ll = long long;
const ll mod = 7 + 1e9;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

void sol()
{
    ll n;
    cin >> n;
    ll ans = log2(n + 1);
    cout << ans << '\n';
}
signed main()
{
    int t = 1;
    cin >> t;
    while (t--)
        sol();
    return 0;
}

C

先从起点出发，然后标记一下可以到达哪里。

接着从终点出发，康康这一行这一列是不是已经之前通过起点到过了。

这就是很直观的思路了。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define pii pair<int, int>
const int N = 1010, mod = 998244353, INF = 0x3f3f3f3f;

char g[N][N];
int vis[N][N];

int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
void solve()
{
    int n, m;
    cin >> n >> m;
    map<int, int> gx, gy;
    int sx, sy, ex, ey;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            cin >> g[i][j];
            if (g[i][j] == 'S')
                sx = i, sy = j;
            if (g[i][j] == 'E')
                ex = i, ey = j;
        }
    }
    queue<pii> q, p;
    q.push({sx, sy});

    while (!q.empty())
    {
        auto [x, y] = q.front();
        q.pop();
        if (vis[x][y])
            continue;
        vis[x][y] = 1;
        for (int i = -1; i <= 1; i++)
        {
            gx[x + i] = gy[y + i] = 1;
        }
        for (int i = 0; i < 4; i++)
        {
            int u = x + dx[i], v = y + dy[i];
            if (u < 0 || v < 0 || u == n || v == m || g[u][v] == '#' || vis[u][v])
                continue;
            q.push({u, v});
        }
    }
    p.push({ex, ey});

    while (!p.empty())
    {
        auto [x, y] = p.front();
        p.pop();
        // cout << x << " " << y << endl;
        if (gx[x] || gy[y])
        {
            cout << "YES\n";
            return;
        }
        if (vis[x][y])
            continue;
        vis[x][y] = 1;

        for (int i = 0; i < 4; i++)
        {
            int u = x + dx[i], v = y + dy[i];
            if (u < 0 || v < 0 || u == n || v == m || g[u][v] == '#' || vis[u][v])
                continue;
            p.push({u, v});
        }
    }
    cout << "NO\n";
}
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0);
    cout.tie(0);
    int T = 1;
    // cin >> T;
    while (T--)
        solve();

    return 0;
}

D

找到质数绝对满足条件。

考察：欧拉筛+哈希。

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#include <bits/stdc++.h>

using namespace std;
#define int long long
using pii = pair<int, int>;
const int N = 1e7 + 10;
int vis[N], primes[N], cnt;
signed main()
{
    for (int i = 2; i < N; i++)
    {
        if (!vis[i])
            primes[cnt++] = i;
        for (int j = 0; primes[j] * i < N; j++)
        {
            vis[primes[j] * i] = 1;
            if (i % primes[j] == 0)
                break;
        }
    }
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        vector<int> a(n);
        unordered_map<int, int> mp;
        for (auto &x : a)
            cin >> x, mp[x] = 1;
        for (int i = 0; i <= 2e5 + 10; i++)
        {
            if (!mp[primes[i]])
            {
                cout << primes[i] << endl;
                break;
            }
        }
    }
    return 0;
}

E

找到最大的m个块。

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#include <iostream>
#include <bits/stdc++.h>
#include <set>
#include <map>
using namespace std;

#define endl "\n"
#define ll long long

void solve()
{
    ll n, m;
    cin >> n >> m;
    vector<pair<ll, ll>> vp(n);
    for (ll i = 0; i < n; i++)
        cin >> vp[i].first;
    for (ll i = 0; i < n; i++)
        cin >> vp[i].second;
    for (ll i = 0; i < n; i++)
        swap(vp[i].first, vp[i].second);
    sort(vp.begin(), vp.end());
    vector<ll> k;
    ll p = 0;
    while (p < n)
    {
        ll rp = vp[p].first + vp[p].second;
        ll sz = 1;
        while (p + 1 < n && rp >= vp[p + 1].first)
        {
            p++;
            sz++;
            rp = max(rp, vp[p].first + vp[p].second);
        }
        p++;
        k.push_back(sz);
    }
    sort(k.begin(), k.end());
    reverse(k.begin(), k.end());
    ll ans = 0;
    for (ll i = 0; i < k.size() && i < m; i++)
    {
        ans += k[i];
    }
    cout << ans << endl;
}

signed main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    ll t;
    cin >> t;
    while (t--)
        solve();
}