牛客小白月赛87

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 312024-02-058.牛客周赛 Round 322024-02-129.2024牛客寒假算法基础集训营12024-02-12

10.牛客小白月赛872024-02-17

11.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

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牛客小白月赛87

小苯的石子游戏

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<int> a(n + 1);
    int x = 0;
    int y = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    int cur = 1;
    for (int i = n; i; i--)
    {
        if (cur & 1)
        {
            x += a[i];
        }
        else
        {
            y += a[i];
        }
        cur++;
    }
    if (x > y)
    {
        puts("Alice");
    }
    else
    {
        puts("Bob");
    }
}

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

小苯的排序疑惑

解题思路：

最小在首位或最大在尾部即可。

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;
    vector<int> a(n + 1);
    int maxa = -1e9 - 1;
    int mina = 1e9 + 1;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        maxa = max(maxa, a[i]);
        mina = min(mina, a[i]);
    }
    if (a[1] == mina || a[n] == maxa)
    {
        puts("YES");
    }
    else
    {
        puts("NO");
    }
}

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

小苯的IDE括号问题（easy）

解题思路:

双向链表删改即可。

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, q;
    cin >> n >> q;
    string s;
    cin >> s;
    int cur = 0;
    n = s.size();
    s = ' ' + s;
    vector<int> l(n + 1, 0), r(n + 1, n + 1);
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == 'I')
        {
            cur = i;
        }
        l[i] = i - 1;
        r[i] = i + 1;
    }
    auto del = [&](int u)
    {
        if (l[u] > 0)
        {
            r[l[u]] = r[u];
        }
        if (r[u] <= n)
        {
            l[r[u]] = l[u];
        }
    };
    while (q--)
    {
        string t;
        cin >> t;
        if (t[0] == 'b')
        {
            array<bool, 2> vis;
            if (r[cur] <= n)
            {
                vis[0] = true;
            }
            if (l[cur] > 0)
            {
                vis[1] = true;
            }
            if (!vis[1])
            {
                continue;
            }
            else
            {
                if (vis[0] && s[l[cur]] == '(' && s[r[cur]] == ')')
                {
                    del(l[cur]);
                    del(r[cur]);
                }
                else
                {
                    del(l[cur]);
                }
            }
        }
        else
        {
            array<bool, 2> vis;
            if (r[cur] <= n)
            {
                vis[0] = true;
            }
            if (vis[0])
            {
                del(r[cur]);
            }
        }
    }
    string ls, rs;
    int t = cur;
    while (true)
    {
        ls += s[t];
        t = l[t];
        if (t == 0)
        {
            break;
        }
    }
    reverse(ls.begin(), ls.end());
    ls.pop_back();
    t = cur;
    while (true)
    {
        rs += s[t];
        t = r[t];
        if (t == n + 1)
        {
            break;
        }
    }
    ls += rs;
    cout << ls << endl;
}

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

小苯的IDE括号问题（hard）

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, q;
    cin >> n >> q;
    string s;
    cin >> s;
    int cur = 0;
    n = s.size();
    s = ' ' + s;
    vector<int> l(n + 1, 0), r(n + 1, n + 1);
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == 'I')
        {
            cur = i;
        }
        l[i] = i - 1;
        r[i] = i + 1;
    }
    auto del = [&](int u)
    {
        if (l[u] > 0)
        {
            r[l[u]] = r[u];
        }
        if (r[u] <= n)
        {
            l[r[u]] = l[u];
        }
    };
    while (q--)
    {
        string t;
        cin >> t;
        if (t[0] == 'b')
        {
            array<bool, 2> vis;
            if (r[cur] <= n)
            {
                vis[0] = true;
            }
            if (l[cur] > 0)
            {
                vis[1] = true;
            }
            if (!vis[1])
            {
                continue;
            }
            else
            {
                if (vis[0] && s[l[cur]] == '(' && s[r[cur]] == ')')
                {
                    del(l[cur]);
                    del(r[cur]);
                }
                else
                {
                    del(l[cur]);
                }
            }
        }
        else if (t[0] == 'd')
        {
            array<bool, 2> vis;
            if (r[cur] <= n)
            {
                vis[0] = true;
            }
            if (vis[0])
            {
                del(r[cur]);
            }
        }
        else if (t[0] == '<')
        {
            if (l[cur] > 0)
            {
                swap(s[cur], s[l[cur]]);
                cur = l[cur];
            }
        }
        else
        {
            if (r[cur] <= n)
            {
                swap(s[cur], s[r[cur]]);
                cur = r[cur];
            }
        }
    }
    string ls, rs;
    int t = cur;
    while (true)
    {
        ls += s[t];
        t = l[t];
        if (t == 0)
        {
            break;
        }
    }
    reverse(ls.begin(), ls.end());
    ls.pop_back();
    t = cur;
    while (true)
    {
        rs += s[t];
        t = r[t];
        if (t == n + 1)
        {
            break;
        }
    }
    ls += rs;
    cout << ls << endl;
}

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

小苯的数组构造

解题思路:

整个$b$数组同时加减一个数不影响答案正确性，我们先默认$b_1=0$。

若$a_i>a_j,(i<j)$，$dif=abs(a_i-a_j)$，我们有两种方法$a_i-dif,a_j+dif$。两种操作对于首位$0$带来的极差影响是一样的，我们往后统一只加不减。

所以，每个元素只需要加到和前缀最大值一样即可。

代码:

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

void solve()
{
    int n;
    cin >> n;
    vector<long long> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    long long pre = a[1];
    long long ans = 0;
    vector<long long> b(n + 1, 0);
    for (int i = 2; i <= n; i++)
    {
        if (a[i] < pre)
        {
            b[i] = pre - a[i];
        }
        pre = max(pre, a[i]);
    }
    for (int i = 1; i <= n; i++)
    {
        cout << b[i] << ' ';
    }
    cout << endl;
}


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

小苯的数组切分

解题思路:

$\mathrm{\&}$:由于是最后一段数组，该操作包含元素越多，区间总权值只可能越小，所以，最后一段只有$a_n$。

接下来就是两段区间和$(1,i)+(i+1,n-1)$，预处理前后缀和后，枚举分界点即可。

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;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    vector<int> pre(n + 1), suf(n + 1);
    for (int i = 1; i <= n; i++)
    {
        pre[i] = pre[i - 1] ^ a[i];
    }
    for (int i = n - 1; i > 0; i--)
    {
        suf[i] = suf[i + 1] | a[i];
    }
    ll ans = 0;
    for (int i = 1; i <= n - 2; i++)
    {
        ans = max(ans, (ll)pre[i] + suf[i + 1] + a[n]);
    }
    cout << ans << endl;
}

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

小苯的逆序对

解题思路:

$g[x]:gcd\mathrm{为}x\mathrm{的}\mathrm{倍}\mathrm{数}\mathrm{的}\mathrm{数}\mathrm{组}\mathrm{对}\mathrm{数}$

$f[x]:gcd\mathrm{恰}\mathrm{好}\mathrm{为}x\mathrm{的}\mathrm{数}\mathrm{组}\mathrm{对}\mathrm{数}$

$f[x]=g[x]-g[2x]-g[3x]-...-g[kx],ps:(k+1)x>n$。

$\lfloor \frac{n}{i}\rfloor :1\sim n\mathrm{内}i\mathrm{的}\mathrm{倍}\mathrm{数}\mathrm{有}\mathrm{多}\mathrm{少}\mathrm{个}$

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;
ll f[N];
ll tr[N];
ll n;

void add(int x, int v)
{
    for (int i = x; i <= n; i += (i & -i))
    {
        tr[i] += v;
    }
}

ll query(int x)
{
    ll res = 0;
    for (int i = x; i; i -= (i & -i))
    {
        res += tr[i];
    }
    return res;
}

void solve()
{
    cin >> n;
    vector<int> pos(n + 1);
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        pos[x] = i;
    }
    for (int i = n; i; i--)
    {
        for (int j = i; j <= n; j += i)
        {
            f[i] += (j / i) - 1 - query(pos[j]);
            add(pos[j], 1);
        }
        for (int j = i; j <= n; j += i)
        {
            add(pos[j], -1);
            if (j != i)
            {
                f[i] -= f[j];
            }
        }
    }
    cout << f[1] << endl;
}

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