牛客周赛 Round 32

合集 - 牛客竞赛(15)

1.牛客小白月赛792023-10-212.牛客小白月赛812023-11-233.牛客小白月赛852024-01-124.牛客小白月赛862024-01-195.牛客周赛 Round 292024-01-256.牛客练习赛1212024-01-267.牛客周赛 Round 312024-02-05

8.牛客周赛 Round 322024-02-12

9.2024牛客寒假算法基础集训营12024-02-1210.牛客小白月赛872024-02-1711.牛客周赛 Round 342024-02-2512.牛客练习赛1222024-03-0313.牛客周赛 Round 352024-03-0314.牛客小白月赛882024-03-0815.牛客周赛49 F 嘤嘤不想找最小喵2024-07-01

收起

牛客周赛 Round 32

小红的 01 背包

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 a, b, c;
    cin >> a >> b >> c;
    cout << a / b * c << endl;
}

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

    return 0;
}

小红的 dfs

解题思路:

观察可知，一共只有三种情况。

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()
{
    vector<string> g(4);
    for (int i = 1; i <= 3; i++)
    {
        cin >> g[i];
        g[i] = ' ' + g[i];
    }
    int a = 0;
    int b = 0;
    int c = 0;
    if (g[1][1] != 'd')
    {
        a++;
    }
    if (g[1][2] != 'f')
    {
        a++;
    }
    if (g[1][3] != 's')
    {
        a++;
    }
    if (g[2][1] != 'f')
    {
        a++;
    }
    if (g[3][1] != 's')
    {
        a++;
    }
    if (g[2][1] != 'd')
    {
        b++;
    }
    if (g[2][2] != 'f')
    {
        b++;
    }
    if (g[2][3] != 's')
    {
        b++;
    }
    if (g[1][2] != 'd')
    {
        b++;
    }
    if (g[3][2] != 's')
    {
        b++;
    }
    if (g[3][1] != 'd')
    {
        c++;
    }
    if (g[3][2] != 'f')
    {
        c++;
    }
    if (g[3][3] != 's')
    {
        c++;
    }
    if (g[1][3] != 'd')
    {
        c++;
    }
    if (g[2][3] != 'f')
    {
        c++;
    }
    int ans = min(a, min(b, c));
    cout << ans << endl;
}

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

    return 0;
}

小红的排列生成

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];
    }
    sort(a.begin() + 1, a.end());
    ll ans = 0;
    for (int i = 1; i <= n; i++)
    {
        ans += abs(a[i] - i);
    }
    cout << ans << endl;
}

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

    return 0;
}

小红的二进制树

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 int N = 1e5 + 10;
vector<int> e[N];
int w[N];
int f[N];

void dfs(int u, int fa)
{
    f[u] = w[u];
    for (auto v : e[u])
    {
        if (v == fa)
        {
            continue;
        }
        dfs(v, u);
        f[u] += f[v];
    }
}

void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    for (int i = 0; i < n; i++)
    {
        w[i + 1] = s[i] - '0';
    }
    for (int i = 1; i < n; i++)
    {
        int a, b;
        cin >> a >> b;
        e[a].push_back(b);
    }
    dfs(1, -1);
    for (int i = 1; i <= n; i++)
    {
        cout << f[i] - w[i] << endl;
    }
}

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

    return 0;
}

小红的回文数

解题思路:

回文数特性：最多只有一个数字出现个数是奇数。

$0\sim 9\mathrm{共}10\mathrm{个}\mathrm{数}\mathrm{字}\mathrm{在}\mathrm{每}\mathrm{个}\mathrm{数}\mathrm{位}\mathrm{上}\mathrm{如}\mathrm{果}\mathrm{是}\mathrm{奇}\mathrm{数}\mathrm{就}\mathrm{是}1，\mathrm{否}\mathrm{则}\mathrm{就}\mathrm{是}0，\mathrm{压}\mathrm{缩}\mathrm{组}\mathrm{成}\mathrm{的}10\mathrm{进}\mathrm{制}\mathrm{数}。i\in [0,2^{10}])$

$dp[i]:\mathrm{前}\mathrm{缀}\mathrm{中}\mathrm{为}i\mathrm{的}\mathrm{状}\mathrm{态}\mathrm{有}\mathrm{多}\mathrm{少}\mathrm{个}$

对于当前状态$cur$，如果前缀和状态也为$cur$，说明从该前缀到当前为止中，所有数字的个数都是偶数个。

对于当前状态$cur$，如果前缀和状态与当前状态只有一个数位不同，说明从该前缀到当前为止中，只有一个数字的个数是奇数个。

上述两种状况能与当前状态构成好整数。

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>>;
const int N = 1ll << 11;
int dp[N];
void solve()
{
    string s;
    cin >> s;
    int cur = 0;
    dp[cur] = 1;
    ll ans = 0;
    for (int i = 0; i < s.size(); i++)
    {
        cur ^= 1 << (s[i] - '0');
        ans += dp[cur];
        for (int j = 0; j < 10; j++)
        {
            int t = cur;
            t ^= 1 << j;
            ans += dp[t];
        }
        dp[cur]++;
    }
    cout << ans << endl;
}

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

    return 0;
}

小红的矩阵修改

解题思路:

状态压缩$dp$。

$dp[i][j]:\mathrm{考}\mathrm{虑}\mathrm{前}i\mathrm{列}，\mathrm{第}i\mathrm{列}\mathrm{为}j\mathrm{时}\mathrm{的}\mathrm{合}\mathrm{法}\mathrm{方}\mathrm{案}\mathrm{的}\mathrm{最}\mathrm{小}\mathrm{操}\mathrm{作}\mathrm{步}\mathrm{数}$。

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>>;
const int N = 1010;
int dp[N][100];
int v[5][N];
bool valid[100];
int n, m;
int b[10];
bool check(int x)
{
    for (int i = 1; i <= n; i++, x /= 3)
    {
        b[i] = x % 3;
    }
    for (int i = 2; i <= n; i++)
    {
        if (b[i] == b[i - 1])
        {
            return false;
        }
    }
    return true;
}

bool check(int a, int b)
{
    for (int i = 1; i <= n; i++)
    {
        if (a % 3 == b % 3)
        {
            return false;
        }
        a /= 3;
        b /= 3;
    }
    return true;
}

void solve()
{
    memset(dp, 0x3f, sizeof dp);
    cin >> n >> m;
    int s = 1;
    for (int i = 1; i <= n; i++)
    {
        s *= 3;
    }
    vector<string> g(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> g[i];
        g[i] = ' ' + g[i];
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (g[i][j] == 'r')
            {
                v[i][j] = 0;
            }
            else if (g[i][j] == 'e')
            {
                v[i][j] = 1;
            }
            else
            {
                v[i][j] = 2;
            }
        }
    }
    ll ans = 1e18;
    for (int i = 1; i <= m; i++)
    {
        if (i == 1)
        {
            for (int j = 0; j < s; j++)
            {
                if (check(j))
                {
                    int cnt = 0;
                    for (int k = 1; k <= n; k++)
                    {
                        if (b[k] != v[k][i])
                        {
                            cnt++;
                        }
                    }
                    dp[i][j] = min(dp[i][j], cnt);
                }
            }
        }
        else
        {
            for (int j = 0; j < s; j++)
            {
                if (check(j))
                {
                    int cnt = 0;
                    for (int k = 1; k <= n; k++)
                    {
                        if (b[k] != v[k][i])
                        {
                            cnt++;
                        }
                    }
                    for (int k = 0; k < s; k++)
                    {
                        if (check(j, k))
                        {
                            dp[i][j] = min(dp[i][j], dp[i - 1][k] + cnt);
                        }
                    }
                }
            }
        }
    }
    for (int i = 0; i < s; i++)
    {
        ans = min(ans, (ll)dp[m][i]);
    }
    cout << ans << endl;
}

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

    return 0;
}