A：空间

256 * 1024 * 1024 / 32 ？？？？！出考场差点没把桌子掀了。

#include <iostream>

using namespace std;

int main() {
    cout << 256 * 1024 * 1024 / 4 << endl;
    return 0;
}

答案：67108864

B:卡片

#include <iostream>

using namespace std;

int main() {
    int a[10];
    for (int i = 0; i <= 9; i++) a[i] = 2021;
    for (int i = 1; i <= 20210; i++) {
        int temp = i;
        bool flag = true;
        while (temp) {
            if (a[temp % 10] == 0) {
                flag = false;
                break;
            }
            a[temp % 10]--;
            temp /= 10;
        }
        if (!flag) {
            cout << i - 1 << endl;
            break;
        }
    }

    return 0;
}

答案：3181

C：直线

直接用set<double, double>存k,b就炸了，会少很多，但是这种存所有直线，所以用判断两个数的误差的方法来判断两个数是否相等，学到了。

#include <iostream>
#include <set>
#include <cmath>
#include <algorithm>

using namespace std;

const int N = 2e5 + 10;

const double eps = 1e-8;    

int cnt = 0;

struct Node {
    double k, b;
    bool operator< (const Node &W) const {
        if (k == W.k) return b < W.b;
        return k < W.k;
    }
} l[N];

int main() {
    for (int x1 = 0; x1 <= 19; x1++) {
        for (int y1 = 0; y1 <= 20; y1++) {
            for (int x2 = 0; x2 <= 19; x2++) {
                for (int y2 = 0; y2 <= 20; y2++) {
                    if (x1 != x2) {
                        double k = (double) (y2 - y1) / (x2 - x1);
                        double b = (double) y1 - k * x1;
                        l[cnt++] = {k, b};   
                    }
                }
            }
        }
    }

    sort(l, l + cnt);

    int res = 1;
    for (int i = 1; i < cnt; i++) {
        if (fabs(l[i].k - l[i - 1].k) > eps || fabs(l[i].b - l[i - 1].b) > eps) res++;
    }

    cout << res + 20 << endl;

    return 0;
}

答案：40257

D：货物摆放

首先用计算器算一下，这个数不是一个平方数，所以暴力分解所有约数即可。

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

typedef long long LL;

int main() {
    LL n = 2021041820210418;
    vector<LL> all;
    for (LL i = 1; i <= n / i; i++) {
        if (n % i == 0) all.push_back(i); 
    }
    int cnt = all.size();
    for (int i = 0; i < cnt; i++) all.push_back(n / all[i]);

    int res = 0;
    for (auto i1 : all) {
        for (auto i2 : all) {
            for (auto i3 : all) {
                if (i1 * i2 * i3 == n) res++;
            }
        }
    }

    cout << res << endl;

    return 0;
}

答案：2430

E：路径

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>

using namespace std;

const int N = 2E6 + 10;

typedef pair<int, int> PII;

int h[3000], e[N], ne[N], w[N], idx;
bool st[3000];
int dist[3000];

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

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

int lcm(int _gcd, int a, int b) {
    return _gcd * (a / _gcd) * (b / _gcd);
}

void dijkstra() {
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, 1});

    while (heap.size()) {
        auto t = heap.top();
        heap.pop();

        int ver = t.second;
        if (st[ver]) continue;
        st[ver] = true;

        for (int i = h[ver]; i != -1; i = ne[i]) {
            int j = e[i];
            if (dist[j] > dist[ver] + w[i]) {
                dist[j] = dist[ver] + w[i];
                heap.push({dist[j], j});
            }
        }
    }
 }

int main() {
    memset(h, -1, sizeof h);
    for (int a = 1; a <= 2021; a++) {
        for (int b = 1; b <= 2021; b++) {
            if (abs(a - b) > 21) continue;
            int _lcm = lcm(gcd(a, b), a, b);
            add(a, b, _lcm), add(b, a, _lcm);
        }
    }

    dijkstra();

    cout << dist[2021] << endl;

    return 0;
}

答案：10266837

F：时间显示

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>

using namespace std;

typedef long long LL;

int main() {
    LL n;
    scanf("%lld", &n);
    LL allTs = n / 1000;
    LL h = allTs / 3600 % 24;
    allTs %= 3600;
    LL m = allTs / 60;
    allTs %= 60;

    printf("%02lld:%02lld:%02lld", h, m, allTs);

    return 0;
}

G:砝码称重

就是个01背包问题，考场写爆搜应该得了还不到一半分。
f[i][j]表示的就是从前i个物品中选，总重量为k的集合。
对这里来说，因为有负数，而我们又不能用负坐标，那么就只好加一个偏移量来求了。

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 110, M = 2e5 + 10;
int n;
bool f[N][M];
int w[N];

int main() {
    cin >> n;
    int m = 0;
    for (int i = 1; i <= n; i++) {
        cin >> w[i];
        m += w[i];
    }
    f[0][m] = true;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 2 * m; j++) {
            f[i][j] |= f[i - 1][j];
            if (j - w[i] >= 0) f[i][j] |= f[i - 1][j - w[i]];
            if (j + w[i] <= 2 * m) f[i][j] |= f[i - 1][j + w[i]];
        }
    }

    int res = 0;
    for (int i = m + 1; i <= m * 2; i++) {
        if (f[n][i]) res++;
    }

    cout << res << endl;

    return 0;
}