2024.1.16做题纪要

硬币

多少有些人类智慧了。。。。。

题解写的还行。

具体就是每次把当前这一位代表的质数 \(i\) 向后每隔 \(i\) 个数除上 \(i\)。

这一位肯定是一个质数，因为若是合数则前面一定会被除上质数。

Kaiserredux

#include <bits/stdc++.h>

long long num[1100000];
long long answer[1100000];

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    freopen("coins.in", "r", stdin);
    freopen("coins.out", "w", stdout);

    for (long long i = 1; i <= 1e6; ++ i) {
        answer[i] = num[i] = 1ll * i * i + 1ll;

    }
    for (long long i = 1; i <= 1000000; ++ i) {
        if (num[i] == 1)
            continue;
        long long x = num[i];
        for (long long j = i; j <= 1000000; j += x) {
            answer[j] = std::min(answer[j], x);
            while (num[j] % x == 0)
                num[j] /= x;
        }
    }
    long long Q, N;
    std::cin >> Q;
    while (Q --) {
        std::cin >> N;
        if (answer[N] == N * N + 1ll) 
            std::cout << -1 << '\n';
        else
            std::cout << answer[N] << ' ' << (N * N + 1) / answer[N] << '\n';
    }
    return 0;
}

猜数

逆天单调队列优化dp + 大眼观察法。

根本不会证QAQ。

56 Road

// ubsan: undefined
// accoders
#include <bits/stdc++.h>

int N;
long long sum[52100], dp[4000][4000];

class Monotonic_Queue {
public:
    int l, r;
    std::pair<int, long long> value[52100];

    Monotonic_Queue() {
        l = 1, r = 0;
    }

    void pop_from_range(int rRange) {
        while (l <= r && value[l].first > rRange)
            l ++;
    }
    
    void emplace(long long val, int pos) {
        while (l <= r && value[r].second > val)
            r --;
        value[++ r] = std::make_pair(pos, val); 
    }

    long long front() {
        return value[l].second;
    }

    int front_first() {
        return value[l].first;
    }

    void clear() {
        l = 1, r = 0;
    }
}queue;

int to[51000];

int Len(int l, int r) {
    return r - l + 1;
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    freopen("guess.in", "r", stdin);
    freopen("guess.out", "w", stdout);

    int S = 3750;
    std::cin >> N;
    for (int r = S + 110; r >= 1; -- r) 
        for (int l = 2; l <= S; ++ l)
            dp[r][l] = 1e18;
    for (int i = 1; i <= N; ++ i)
        to[i] = i % 3800, sum[i] = 1e18;
    sum[1] = 0;
    long long answer = 0;
    for (int r = 2; r <= N; ++ r) {
        int mid = r;
        queue.clear();
        for (int l = r - 1; r - l + 1 <= std::min(r, S); -- l) {
            queue.emplace(dp[to[r]][Len(l + 1, r)] + l, l);
            while (mid > l && dp[to[mid - 1]][Len(l, mid - 1)] >= dp[to[r]][Len(mid + 1, r)]) {
                mid --;
                queue.pop_from_range(mid);
            }
            dp[to[r]][Len(l, r)] = std::min(queue.front(), dp[to[mid]][Len(l, mid)] + mid + 1);
            sum[r] = std::min(sum[r], std::max(sum[l - 1], dp[to[r]][Len(l + 1, r)]) + l);
        }
        answer += sum[r];
    }
    std::cout << answer << '\n';
    return 0;
}

P3403 跳楼机

同余最短路板子题。

我们先设 \(f_i\) 表示我们只用操作 \(2,3\) 使得 \(f_i \% x = i\)，并且让 \(f_i\) 最小。每次将 \(i\) 连向 \((i+y)\%x\) 和 \((i+z)\%x\)，边权分别为 \(y\) 和 \(z\)。然后跑最短路就行了。

燃烧的世界

#include <bits/stdc++.h>

typedef long long ll;

class Point {
public:
    ll val, poi;

    bool operator <(const Point &b) const {
        return val > b.val;
    }

    Point(ll _val, ll _poi) {
        val = _val;
        poi = _poi;
    }
};

int cnt, head[110000], next[210000], to[210000];
ll value[210000];

void AddEdge(int u, int v, ll val) {
    ++ cnt;
    next[cnt] = head[u];
    head[u] = cnt;
    to[cnt] = v;
    value[cnt] = val;
}

ll f[110000];
bool visit[110000];
ll H, X, Y, Z;

void Dijkstra(int begin) {
    memset(visit, 0, sizeof(visit));
    for (int i = 0; i < X; ++ i)
        f[i] = 2e18;
    f[1] = 1;
    std::priority_queue<Point> queue;

    queue.emplace(f[1], 1);

    while (queue.size()) {
        int now = queue.top().poi;
        queue.pop();

        if (visit[now])
            continue;

        visit[now] = true;
        for (int i = head[now]; i; i = next[i]) {
            if (f[to[i]] > f[now] + value[i]) {
                f[to[i]] = f[now] + value[i];
                queue.emplace(f[to[i]], to[i]);
            }
        }
    }
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    std::cin >> H >> X >> Y >> Z;
    if (X == 1 || Y == 1 || Z == 1) {
        std::cout << H << '\n';
        return 0;
    }
    if (X > Y)
        std::swap(X, Y);
    if (X > Z)
        std::swap(X, Z);
    for (int i = 0; i < X; ++ i) {
        AddEdge(i, (i + Y) % X, Y);
        AddEdge(i, (i + Z) % X, Z);
    }
    Dijkstra(1);
    ll answer = 0;
    for (int i = 0; i < X; ++ i) {
        if (f[i] > H)
            continue;
        answer += (H - f[i]) / X + 1;
    }
    std::cout << answer << '\n';
    return 0;
}