牛客练习赛121

合集 - 牛客竞赛(15)

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6.牛客练习赛1212024-01-26

7.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12024-02-1210.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

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牛客练习赛121

出题人题解 | 牛客练习赛 121 题解

小念吹气球

解题思路:

字符长度加字符种类数。

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;

void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    map<char, bool> q;
    for (auto c : s)
    {
        q[c] = true;
    }
    cout << n + q.size() << endl;
}

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

    return 0;
}

You Brought Me A Gentle Breeze on the Field

解题思路:

如果只有一个糖果，必败状态，则小念输。

如果糖果在$2\sim m+1$，那么小念赢，因为一定能一步到必败状态。

否则，能连取的赢。因为不能连取的人，一定无法先进入必败状态。只要能连取的人将连取机会留到$m+2\sim 1+2m$，此时，自己一定能转入必败状态，而对面不行。

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;

void solve()
{
    int n, m, p;
    cin >> n >> m >> p;
    if (n == 1)
    {
        puts("YangQiShaoNian");
        return;
    }
    if (n > 1 + m)
    {
        if (p == 1)
        {
            puts("YangQiShaoNian");
        }
        else
        {
            puts("XiaoNian");
        }
    }
    else
    {
        puts("XiaoNian");
    }
}

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

    return 0;
}

氧气少年的水滴 2

解题思路:

模拟。

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;

void solve()
{
    int n, p;
    cin >> n >> p;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    int i = p - 1;
    int j = p + 1;
    a[p]++;
    int l = 0;
    int r = 0;
    if (a[p] == 10)
    {
        l++;
        r++;
    }
    int cnt = 0;
    while (i > 0 && j <= n)
    {
        cnt++;
        if (cnt > n + 1)
        {
            break;
        }
        if (a[i] + l >= 10)
        {
            l -= 10 - a[i];
            i--;
            l++;
            r++;
        }
        else
        {
            a[i] += l;
            l = 0;
        }
        if (a[j] + r >= 10)
        {
            r -= 10 - a[j];
            j++;
            r++;
            l++;
        }
        else
        {
            a[j] += r;
            r = 0;
        }
    }
    cnt = 0;
    while (i > 0)
    {
        if (a[i] + l >= 10)
        {
            l -= 10 - a[i];
            i--;
            l++;
            r++;
        }
        else
        {
            l = 0;
            break;
        }
    }
    cnt = 0;
    while (j <= n)
    {
        if (a[j] + r >= 10)
        {
            r -= 10 - a[j];
            j++;
            r++;
            l++;
        }
        else
        {
            r = 0;
            break;
        }
    }
    cout << l << ' ' << r << endl;
}

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

    return 0;
}

氧气少年的 LCM

解题思路:

$$\begin{gathered} g=gcd(x,y) \\ x=k_1\ast g \\ y=k_2\ast g \\ lcm(x,y)=\frac{x\ast y}{g}=k_1\ast k_2\ast g=k_2\ast x \end{gathered}$$.

如上，我们只要构建出$k_2\ast x$即可。

通过加法构造出$2^1\ast x,2^2\ast x,..{.2}^{32}\ast x$.

我们将$k_2$进行二进制拆分,然后按位将需要的$2^k\ast x$累加起来即可。

注意点:

1. 由于我们需要数字两两相加，记得每个数字构造两个。
2. 题目明确说了，集合内每个元素不超过${10}^{18}$。（本人这点没注意，卡了近一个小时).

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;

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}

struct option
{
    ll t, a, b;
};

void solve()
{
    ll x, y;
    cin >> x >> y;
    ll tar = lcm(x, y);
    if (tar == x || tar == y)
    {
        puts("0");
    }
    else
    {
        ll g = gcd(x, y);
        vector<option> v;
        v.push_back({1, x, y});
        v.push_back({1, x, y});
        ll k1 = x / g;
        ll k2 = y / g;
        ll cur = g;
        vector<ll> p(40), b(40);
        p[0] = x;
        b[0] = g;
        for (int i = 1; i <= 32; i++)
        {
            if (cur + cur > 1e18)
            {
                break;
            }
            v.push_back({2, cur, cur});
            v.push_back({2, cur, cur});
            cur += cur;
            b[i] = cur;
        }
        bool vis = false;
        for (int i = 32; i >= 0; i--)
        {
            if (k1 >> i & 1)
            {
                if (vis == false)
                {
                    vis = true;
                    cur = b[i];
                }
                else
                {
                    v.push_back({2, b[i], cur});
                    cur += b[i];
                }
            }
        }
        cur = x;
        for (int i = 1; i <= 32; i++)
        {
            if (cur + cur > 1e18)
            {
                break;
            }
            v.push_back({2, cur, cur});
            v.push_back({2, cur, cur});
            cur += cur;
            p[i] = cur;
        }
        cur = 0;
        vis = false;
        for (int i = 32; i >= 0; i--)
        {
            if (k2 >> i & 1)
            {
                if (vis == false)
                {
                    vis = true;
                    cur = p[i];
                }
                else
                {
                    v.push_back({2, p[i], cur});
                    cur += p[i];
                }
            }
        }
        cout << v.size() << endl;
        for (auto t : v)
        {
            cout << t.t << ' ' << t.a << ' ' << t.b << endl;
        }
    }
}

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

    return 0;
}

氧气少年逛超市 3

解题思路:

观察范围，时间复杂度$O(n^2)$可过。

对于前$min(a+b,n)$个物品，我们一定使用券。

并且一定是价格贵的物品先用，减额力度大的券。

$f[i][j]:\mathrm{考}\mathrm{虑}\mathrm{前}i\mathrm{个}\mathrm{物}\mathrm{品}，\mathrm{我}\mathrm{们}\mathrm{用}\mathrm{了}j\mathrm{个}\mathrm{折}\mathrm{扣}\mathrm{卷}。$

$$f[i][j]=\{\begin{array}{l}f[i-1][j-1]+a[i]\times \frac{x[j]}{100.0}\\ f[i-1][j]+a[i]-y[i-j]\end{array}$$

注意转移限制区间条件。

$$ans=min(f[min(a+b,n)][0,1,2,...,a])+\sum _{i=a+b+1}^n{p}_i$$

解释，为什么答案统计前面是$f[min(a+b,n)][0,1,2,...,a]$而不是$f[min(a+b,n)][a]$。

因为$0\le a+b\le 2n$,不一定所有的券都会用上。

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;
const int N = 5010;
double f[N][N];

void solve()
{
    int n;
    cin >> n;
    vector<double> a(n + 1), x(1), y(1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i <= 2; i++)
    {
        int m;
        cin >> m;
        for (int j = 1; j <= m; j++)
        {
            double t;
            cin >> t;
            if (i == 1)
            {
                x.push_back(t);
            }
            else
            {
                y.push_back(t);
            }
        }
    }
    sort(a.begin() + 1, a.end());
    reverse(a.begin() + 1, a.end());
    sort(x.begin() + 1, x.end());
    sort(y.begin() + 1, y.end());
    reverse(y.begin() + 1, y.end());
    double ans = 1e18;
    double cur = 0;
    for (int i = 1; i <= n; i++)
    {
        cur += a[i];
        for (int j = 0; j <= n; j++)
        {
            f[i][j] = cur;
        }
    }
    for (int i = 0; i <= n; i++)
    {
        f[0][i] = 0;
    }
    int m = min(n, (int)x.size() + (int)y.size() - 2);
    for (int i = 1; i <= m; i++)
    {
        for (int j = 0; j <= x.size() - 1; j++)
        {
            if (j > 0)
            {
                f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] * x[j] / 100.0));
            }
            if (i - j > 0 && i - j < y.size())
            {
                f[i][j] = min(f[i][j], f[i - 1][j] + max(0.0, a[i] - y[i - j]));
            }
        }
    }
    // cout << f[1][0] << endl;
    for (int i = 0; i <= x.size() - 1; i++)
    {
        ans = min(ans, f[m][i]);
    }
    for (int i = m + 1; i <= n; i++)
    {
        ans += a[i];
    }
    printf("%.14lf\n", ans);
}

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

    return 0;
}