牛客周赛 Round 65（A—G）

比赛链接

牛客周赛 Round 65

A题

思路

谁的单价低就全选哪一个。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 2e5 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, a, b;
void solve()
{
    cin >> n >> a >> b;
    if (a <= b)
    {
        cout << n / a << endl;
    }
    else
        cout << n / b << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

B题

思路

暴力枚举。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e3 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, m;
char s[N][N];
bool rain(int x, int y)
{
    return (x >= 1 && x <= n && y >= 1 && y <= m && s[x][y] == '*');
}
bool check(int x, int y)
{
    int cnt = 0;
    if (rain(x, y))
        cnt++;
    if (rain(x, y + 1))
        cnt++;
    if (rain(x + 1, y))
        cnt++;
    if (rain(x + 1, y + 1))
        cnt++;
    return cnt == 4;
}
void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> s[i][j];
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (check(i, j))
                ans++;
        }
    }
    cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

C题

思路

从大到小排序，之后交换最大值和最小值的位置，遍历一遍比较谁大谁小即可。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e3 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;

int n;
int a[N];
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    sort(a + 1, a + 1 + n, [&](int x, int y)
         { return x > y; });
    swap(a[1], a[n]);
    int op1 = 0, op2 = 0;
    for (int i = 1; i <= n; i++)
    {
        if (i & 1)
            op1 += a[i];
        else
            op2 += a[i];
    }
    if (op1 > op2)
    {
        cout << "kou" << endl;
        return;
    }
    if (op1 == op2)
    {
        cout << "draw" << endl;
        return;
    }
    cout << "yukari" << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

D题

思路

$k$最大为$10$，我们考虑使用二进制状压。

对于第$i$个人，我们将第$j$个药物能够治疗他的症状情况当作一个二进制数（即当病人的第$k$位为$1$时，才会保留药物的第$k$位的$1$的值）。

枚举每一个药物选或不选的情况。即二进制数中，第$k$位为$1$，则代表使用第$k$个药物，否则代表不使用第$k$个药物。如果最后的值$or$的结果刚好为病人症状的二进制数，则用当前使用药物的数量更新答案。

时间复杂度：$O(2^{k}nk)$。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e4 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, m, k;
int a[N][25], b[15][25];
string s;
int qmi(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = res * a;
        b >>= 1;
        a = a * a;
    }
    return res;
}
void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> s;
        for (int j = 0; j < m; j++)
        {
            if (s[j] == '1')
                a[i][j + 1] = 1;
            else
                a[i][j + 1] = 0;
        }
    }
    cin >> k;
    for (int i = 1; i <= k; i++)
    {
        cin >> s;
        for (int j = 0; j < m; j++)
        {
            if (s[j] == '1')
                b[i][j + 1] = 1;
            else
                b[i][j + 1] = 0;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        vector<int> v;
        for (int j = 1; j <= k; j++)
        {
            int res = 0;
            for (int o = 1; o <= m; o++)
            {
                if (a[i][o] == 1 && b[j][o] == 1)
                {
                    res += (qmi(2, o - 1));
                }
            }
            v.push_back(res);
        }
        int sum = 0;
        for (int j = 1; j <= m; j++)
        {
            if (a[i][j] == 1)
                sum += qmi(2, j - 1);
        }
        int minn = inf;
        for (int val = 0; val < qmi(2, k); val++)
        {
            int res = 0, cnt = 0;
            for (int j = 0; j < k; j++)
            {
                int bit = (val >> j) & 1;
                if (bit)
                {
                    cnt++;
                    res |= v[j];
                }
            }
            if (res == sum)
            {
                minn = min(minn, cnt);
            }
        }
        if (minn == inf)
            minn = -1;
        cout << minn << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

E题

思路

如果要求最小值：如果$a[i]$为未知气温，则令$a[i]=a[i-1]-x+1$，与温度的下限取$max$。

如果要求最大值：如果$a[i]$为未知气温，则令$a[i]=a[i-1]-x$，与温度的下限取$max$。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e2 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, x;
int a[N], b[N];
void solve()
{
    cin >> n >> x;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        b[i] = a[i];
    }
    int ans = 0;
    // 求最大
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
        {
            if (a[i] == -999)
                a[i] = 50;
            continue;
        }
        if (a[i] != -999)
        {
            if (a[i - 1] - a[i] >= x)
                ans++;
        }
        else
        {
            if (a[i - 1] - x >= -50)
            {
                a[i] = a[i - 1] - x;
                ans++;
            }
            else
            {
                a[i] = 50;
            }
        }
    }
    cout << ans << " ";
    ans = 0;
    // 求最小
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
        {
            if (b[i] == -999)
                b[i] = -50;
            continue;
        }
        if (b[i] != -999)
        {
            if (b[i - 1] - b[i] >= x)
                ans++;
        }
        else
        {
            b[i] = max(-50ll, b[i - 1] - x + 1);
        }
    }
    cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

F题

思路

原理同$E$题，只是改一下数据范围即可。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e5 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, x;
int a[N], b[N];
void solve()
{
    cin >> n >> x;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        b[i] = a[i];
    }
    int ans = 0;
    // 求最大
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
        {
            if (a[i] == -999999999)
                a[i] = 5e8;
            continue;
        }
        if (a[i] != -999999999)
        {
            if (a[i - 1] - a[i] >= x)
                ans++;
        }
        else
        {
            if (a[i - 1] - x >= -5e8)
            {
                a[i] = a[i - 1] - x;
                ans++;
            }
            else
            {
                a[i] = 5e8;
            }
        }
    }
    cout << ans << " ";
    ans = 0;
    // 求最小
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
        {
            if (b[i] == -999999999)
                b[i] = -5e8;
            continue;
        }
        if (b[i] != -999999999)
        {
            if (b[i - 1] - b[i] >= x)
                ans++;
        }
        else
        {
            b[i] = max(-500000000ll, b[i - 1] - x + 1);
        }
    }
    cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}

G题

思路

我们考虑构造$\mathrm{010101...}$或$\mathrm{101010...}$的基地。

当$n$为偶数时：如果$\frac{n}{2}>m$，则代表连最起码的基地都无法构造，直接输出$-1$。如果$m-\frac{n}{2}$为奇数，则不可能构造成功，因为每一次操作，都要让基地中最小的数加上$2$。

当$n$为奇数时：如果$m-\frac{n}{2}$为偶数，则构造$\mathrm{010101...}$这样的基地，否则就构造$\mathrm{101010...}$这样的基地。剩下的操作同偶数。

代码

#include 
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 1e5 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, m;
void solve()
{
    cin >> n >> m;
    if (n % 2 == 0)
    {
        int op = n / 2;
        if (op > m || abs(m - op) & 1)
        {
            cout << -1 << endl;
            return;
        }
        vector<int> v(n + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            if (i % 2 == 0)
                v[i] = 1;
        }
        m -= (n / 2);
        int num = m / (n * 2);
        for (int i = 1; i <= n; i++)
        {
            v[i] += num * 2;
        }
        m = m - num * n * 2;
        num = m / n;
        if (num == 0)
        {
            for (int i = 1; i <= n; i += 2)
            {
                if (m <= 0)
                    break;
                v[i] += 2;
                m -= 2;
            }
        }
        else
        {
            for (int i = 1; i <= n; i += 2)
            {
                v[i] += num * 2;
            }
            m = m - num * n;
            for (int i = 2; i <= n; i += 2)
            {
                if (m <= 0)
                    break;
                v[i] += 2;
                m -= 2;
            }
        }
        for (int i = 1; i <= n; i++)
        {
            cout << v[i] << " ";
        }
        cout << endl;
    }
    else
    {
        int op = n / 2; // 1的个数

        if (op > m)
        {
            cout << -1 << endl;
            return;
        }
        bool ok = false;
        if ((abs(m - op) & 1)) // 改为1开头
        {
            ok = true;
        }
        if (ok)
            op++;
        if (op > m)
        {
            cout << -1 << endl;
            return;
        }
        vector<int> v(n + 1, 0);
        if (!ok)
        { // 0开头
            for (int i = 1; i <= n; i++)
            {
                if (i % 2 == 0)
                    v[i] = 1;
            }

            m -= (n / 2);
            int num = m / (n * 2);

            for (int i = 1; i <= n; i++)
            {
                v[i] += num * 2;
            }

            m = m - num * n * 2;
            num = m / (((n / 2) + 1) * 2);
            if (num == 0)
            {
                for (int i = 1; i <= n; i += 2)
                {
                    if (m <= 0)
                        break;
                    v[i] += 2;
                    m -= 2;
                }
            }
            else
            {
                for (int i = 1; i <= n; i += 2)
                {
                    v[i] += num * 2;
                }

                m = m - num * (((n / 2) + 1) * 2);
                for (int i = 2; i <= n; i += 2)
                {
                    if (m <= 0)
                        break;
                    v[i] += 2;
                    m -= 2;
                }
            }
        }
        else
        {
            for (int i = 1; i <= n; i++)
            {
                if (i & 1)
                    v[i] = 1;
            }

            m -= ((n / 2) + 1);
            int num = m / (n * 2);

            for (int i = 1; i <= n; i++)
            {
                v[i] += num * 2;
            }

            m = m - num * n * 2;
            num = m / n;
            if (num == 0)
            {
                for (int i = 2; i <= n; i += 2)
                {
                    if (m <= 0)
                        break;
                    v[i] += 2;
                    m -= 2;
                }
            }
            else
            {
                for (int i = 2; i <= n; i += 2)
                {
                    v[i] += num * 2;
                }

                m = m - num * n;
                for (int i = 1; i <= n; i += 2)
                {
                    if (m <= 0)
                        break;
                    v[i] += 2;
                    m -= 2;
                }
            }
        }

        for (int i = 1; i <= n; i++)
        {
            cout << v[i] << " ";
        }
        cout << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}