例如

• $17$ 题按原题面码风第 $15$ 行应为 reverse(d.begin(), d.end());，但洛谷使用的码风第 $15$ 行是 d.push_back(k);
• $18$ 题按原题面码风第 $24$ 行应为h = m;，但洛谷使用的码风第 $24$ 行是 if (f0(a, m, k)) {
• $18$ 题按原题面码风第 $22$ 行应为 int m = g + (h - g) / 2;，但洛谷使用的码风第 $22$ 行是 while (g < h) {

再提个建议：建议把代码每行前面都加上行号（手数二十多行代码是真的痛苦）

附上17题、18题原码风的代码

• 17

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

long long solve1(int n) {
    vector<bool> p(n + 1, true);
    vector<long long> f(n + 1, 0), g(n + 1, 0);
    f[1] = 1;
    for (int i = 2; i * i <= n; i++) {
        if (p[i]) {
            vector<int> d;
            for (int k = i; k <= n; k *= i)	d.push_back(k);
            reverse(d.begin(), d.end());
            for (int k : d) {
                for (int j = k; j <= n; j += k) {
                    if (p[j]) {
                        p[j] = false;
                        f[j] = i;
                        g[j] = k;
                    }
                }
            }
        }
    }
    for (int i = sqrt(n) + 1; i <= n; i++) {
        if (p[i]) {
            f[i] = i;
            g[i] = i;
        }
    }
    long long sum = 1;
    for (int i = 2; i <= n; i++) {
        f[i] = f[i / g[i]] * (g[i] * f[i] - 1) / (f[i] - 1);
        sum += f[i];
    }
    return sum;
}

long long solve2(int n) {
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
        sum += i * (n / i);
    }
    return sum;
}

int main() {
    int n;
    cin >> n;
    cout << solve1(n) << endl;
    cout << solve2(n) << endl;
    return 0;
}

• 18

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

using namespace std;

bool f0(vector<int> &a, int m, int k) {
    int s = 0;
    for (int i = 0, j = 0; i < a.size(); i++) {
        while (a[i] - a[j] > m)	j++;
        s += i - j;
    }
    return s >= k;
}

int f(vector<int> &a, int k) {
    sort(a.begin(), a.end());

    int g = 0;
    int h = a.back() - a[0];
    while (g < h) {
        int m = g + (h - g) / 2;
        if (f0(a, m, k)) {
            h = m;
        } else {
            g = m + 1;
        }
    }

    return g;
}

int main() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n, 0);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cout << f(a, k) << endl;
    return 0;
}