牛客多校第三场补题

A-Clam and Fish
思维模拟

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
double pi = acos(-1);
const double eps = 1e-9;
const int inf = 1e9 + 7;
const int maxn = 5005;

int main()
{
    fastio;
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        string s;
        cin >> s;
        int ans = 0, clam = 0, p = 1;
        s = '0' + s;
        for (int i = 1; i <= n; i++)
        {
            if (s[i] == '2' || s[i] == '3')ans++;
            else if (s[i] == '0' && clam)ans++, clam--;
            else if (s[i] == '1')
            {
                int j = max(i + 1, p);
                for (; j <= n; j++)
                {
                    if (s[j] == '0')
                    {
                        s[j] = '2';
                        p = j + 1;
                        break;
                    }
                }
                if (j > n)
                {
                    if (clam)
                        clam--, ans++;
                    else
                        clam++;
                    p = n + 1;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
 
}

---

---

B-Classical String Problem
思维模拟

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
double pi = acos(-1);
const double eps = 1e-9;
const int inf = 1e9 + 7;
const int maxn = 5005;
 
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return f * x;
}
 
 
int main()
{
    fastio;
    string s;
    cin >> s;
    int t;
    cin >> t;
    int tmp = 0;
    int len = s.length();
    while (t--)
    {
        char o;
        int x;
        cin >> o >> x;
        if (o == 'M')
            tmp += x;
        else
        {
            if (tmp == 0)
                cout << s[x - 1] << endl;
            else if (tmp > 0)
                cout << s[(x + tmp - 1) % len] << endl;
            else
                cout << s[((len + tmp + x - 1) % len + len) % len] << endl;
        }
    }
    return 0;
 
}

---

---

E-Two Matchings

贪心 DP ，打的时候真不知道这是个dp

这样我们就能以4和6一组作为状态进行dp了,最终状态肯定能被拆成数个4+数个6，否则不优

（解释：组与组之间断裂，dp寻找的就是在哪里断裂最优，然后组之间的每个差值只被计入2次答案是下界（最优），无法找到更优的连法)

状态转移方程: (dp[i] = min(dp[i], dp[i - 4] + 2 * (a[i] - a[i - 3])))
(dp[i] = min(dp[i], dp[i - 6] + 2 * (a[i] - a[i - 5]))|i >= 6)

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
double pi = acos(-1);
const double eps = 1e-12;
const int inf = 1e17 + 7;
const int maxn = 1e5 + 10;
 
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return f * x;
}
 
ll dp[maxn];
 
int main()
{
    fastio;
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vector<ll>a(n + 1), dp(n + 1, inf);
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        sort(a.begin() + 1, a.end());
        dp[0] = 0;
        for (int i = 4; i <= n; i++)
        {
            dp[i] = min(dp[i], dp[i - 4] + 2 * (a[i] - a[i - 3]));//容斥
            if (i >= 6)dp[i] = min(dp[i], dp[i - 6] + 2 * (a[i] - a[i - 5]));
        }
        cout << dp[n] << endl;
    }
    return 0;
 
}

F-Fraction Construction Problem

构造 EXGCD 不定方程

打的时候写了个EXGCD解了下ab不互素的不定方程，然后发现不太会整其他的情况，又去想了个sb的错误做法，我只能爬

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
double pi = acos(-1);
const double eps = 1e-12;
const int maxn = 2e6 + 10;
const int inf = 1e9;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return f * x;
}
ll x, y;

ll v[maxn], p[maxn], cnt;

void init_e()
{
    memset(v, 0, sizeof(v));
    cnt = 0;
}

void Euler_sieve(int n)
{
    //init_e();
    cnt = 0;
    for (int i = 2; i <= n; i++) {
        if (v[i] == 0) { v[i] = i; p[++cnt] = i; }
        for (int j = 1; j <= cnt; j++)
        {
            if (p[j] > v[i] || p[j] > n / i)break;
            v[i * p[j]] = p[j];
        }
    }
}

ll exgcd(ll a, ll b, ll& x, ll& y)
{
    if (!b)
    {
        x = 1;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

int gcd(int a, int b)
{
    while (b)
    {
        ll tmp = b;
        b = a % b;
        a = tmp;
    }
    return a;
}


int main()
{
    fastio;
    int t;
    cin >> t;
    Euler_sieve(2e6 + 5);
    while (t--)
    {
        x = y = 0;
        ll a, b;
        cin >> a >> b;
        ll g = gcd(a, b);
        if (g != 1)
        {
            cout << a / g + 1 << " " << b / g << " " << 1 << " " << b / g << endl;
            continue;
        }
        g = 1;
        if (b == 1 || v[b] == b) {
            cout << -1 << " " << -1 << " " << -1 << " " << -1 << endl;
            continue;
        }
        ll k = v[b];
        while (b % k == 0)
            b /= k, g *= k;
        if (b == 1) {
            cout << -1 << " " << -1 << " " << -1 << " " << -1 << endl;
            continue;
        }
        //cout << p << " " << b << endl;
        exgcd(b, g, x, y);
        if (x > 0)
            cout << x * a << " " << g << " " << -y * a << " " << b << endl;
        else
            cout << y * a << " " << b << " " << -x * a << " " << g << endl;
    }
    return 0;

}

最后！！！画了一万年也没有画完的线稿来了，什么时候才能有时间画完啊QAQ