牛客周赛 Round 31

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1212024-01-26

7.牛客周赛 Round 312024-02-05

8.牛客周赛 Round 322024-02-129.2024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 31

小红小紫替换

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()
{
    string s;
    cin >> s;
    if (s == "kou")
    {
        cout << "yukari" << endl;
    }
    else
    {
        cout << s << endl;
    }
}

int main()
{

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

    return 0;
}

小红的因子数

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()
{
    ll x;
    cin >> x;
    int cnt = 0;
    for (int i = 2; i <= x / i; i++)
    {
        if (x % i == 0)
        {
            cnt++;
            while (x % i == 0)
            {
                x /= i;
            }
        }
    }
    if (x > 1)
    {
        cnt++;
    }
    cout << cnt << endl;
}

int main()
{

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

    return 0;
}

小红的字符串中值

解题思路:

对于每一个询问字符，考虑以它为中心能构造多少子串，即尽量往两边扩。

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;
    char c;
    scanf("%d %c", &n, &c);
    string s;
    cin >> s;
    ll ans = 0;
    for (int i = 0; i < n; i++)
    {
        if (s[i] == c)
        {
            int l = i - 0;
            int r = n - 1 - i;
            ans += min(l, r) + 1;
        }
    }
    cout << ans << endl;
}

int main()
{

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

    return 0;
}

小红数组操作

解题思路:

用$map$模拟链表操作。

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 = 2e5 + 10;
map<ll, int> l, r;

void solve()
{
    int q;
    cin >> q;
    map<int, int> v;
    int h = 0;
    int cnt = 0;
    while (q--)
    {
        int t;
        cin >> t;
        if (t == 1)
        {
            int x, y;
            cin >> x >> y;
            cnt++;
            if (y == 0)
            {
                r[x] = h;
                l[h] = x;
                h = x;
            }
            else
            {
                l[x] = y;
                r[x] = r[y];
                r[y] = x;
                l[r[x]] = x;
            }
        }
        else
        {
            int x;
            cin >> x;
            r[l[x]] = r[x];
            l[r[x]] = l[x];
            if (x == h)
            {
                h = r[x];
            }
            cnt--;
        }
    }
    // if (cnt == 0)
    // {
    //     cout << 0 << endl;
    //     return;
    // }
    vector<int> ans;
    while (h)
    {
        ans.push_back(h);
        h = r[h];
    }
    cout << ans.size() << endl;
    for (auto c : ans)
    {
        cout << c << ' ';
    }
}

int main()
{

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

    return 0;
}

小红的子集取反

解题思路:

$dp[i][j]:\mathrm{表}\mathrm{示}\mathrm{取}\mathrm{完}\mathrm{前}i\mathrm{个}\mathrm{数}\mathrm{字}\mathrm{后}，\mathrm{此}\mathrm{时}\mathrm{总}\mathrm{和}\mathrm{为}j\mathrm{的}\mathrm{最}\mathrm{小}\mathrm{取}\mathrm{反}\mathrm{数}。\mathrm{对}\mathrm{于}\mathrm{第}i\mathrm{个}\mathrm{数}\mathrm{字}\mathrm{我}\mathrm{们}\mathrm{如}\mathrm{果}\mathrm{减}\mathrm{去}\mathrm{就}\mathrm{是}\mathrm{取}\mathrm{反}，\mathrm{加}\mathrm{上}\mathrm{就}\mathrm{无}\mathrm{需}\mathrm{增}\mathrm{加}\mathrm{操}\mathrm{作}\mathrm{数}。$

注意值域存在负数，所以我们对初始状态进行偏移。

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 int N = 40000;
const int inf = 1 << 30;
int dp[210][2 * N + 10];

void solve()
{
    int n;
    cin >> n;
    memset(dp, 0x3f, sizeof dp);
    dp[0][40000] = 0;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        for (int j = 0; j <= 2 * N; j++)
        {
            if (j + x <= 2 * N)
            {
                dp[i][j] = min(dp[i][j], dp[i - 1][j + x] + 1);
            }
            if (j - x >= 0)
            {
                dp[i][j] = min(dp[i][j], dp[i - 1][j - x]);
            }
        }
    }
    if (dp[n][N] > n)
    {
        dp[n][N] = -1;
    }
    cout << dp[n][N] << endl;
}

int main()
{

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

    return 0;
}

小红的连续段

解题思路:

字符总长度总是为$x+y$。

我们枚举连续段数量，发现只有两种情况：

• $(a,b,a,b,..)$，即$a$连续段开头，此时$a$段的数量必定为$\lceil \frac{i}{2}\rceil$
• $(b,a,b,a,...)$，即$b$连续段开头,此时$b$段的数量必定为$\lceil \frac{i}{2}\rceil$

我们设$a$段数量为$cnta$,$b$段数量为$cntb$。

将$x$个$a$分为$cnta$个非空集合的方案数为$(\genfrac{}{}{0ex}{}{x-1}{cnta-1})$

对$b$同理。

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;
const int mod = 1e9 + 7;
ll c[N][N];
void solve()
{
    int x, y;
    cin >> x >> y;
    int n = x + y;
    c[0][0] = 1;
    for (int i = 1; i <= 1001; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            if (j == 0)
            {
                c[i][j] = 1;
            }
            else
            {
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        ll ans = 0;
        int ca = i / 2;
        int cb = i - ca;
        if (i < 2)
        {
            cout << 0 << endl;
            continue;
        }
        ll res = 1;
        if (ca <= x)
        {
            res = res * c[x - 1][ca - 1] % mod;
        }
        else
        {
            res = 0;
        }
        if (cb <= y)
        {
            res = res * c[y - 1][cb - 1] % mod;
        }
        else
        {
            res = 0;
        }
        ans += res;
        ans %= mod;
        res = 1;
        swap(ca, cb);
        if (ca <= x)
        {
            res = res * c[x - 1][ca - 1] % mod;
        }
        else
        {
            res = 0;
        }
        if (cb <= y)
        {
            res = res * c[y - 1][cb - 1] % mod;
        }
        else
        {
            res = 0;
        }
        ans += res;
        ans %= mod;
        cout << ans << endl;
    }
}

int main()
{

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

    return 0;
}