L2-017 人以群分
偶数对半分最好，奇数的话外向多一个差就更大
#include <bits/stdc++.h>
using namespace std;
#define int long long
using PII = pair<int, int>;void ylh_() {int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; ++i) {cin >> a[i];}sort(a.begin() + 1, a.end());vector<int> pre(n + 1);for (int i = 1; i <= n; ++i) {pre[i] = pre[i - 1] + a[i];}int diff = pre[n] - 2 * pre[n / 2];cout << "Outgoing #: " << n / 2 + (n % 2 == 1) << '\n';cout << "Introverted #: " << n / 2 << '\n';cout << "Diff = " << diff;
}int32_t main() {ios::sync_with_stdio(0);cin.tie(0);int T = 1;// cin >> T;while (T--) {ylh_();}
}
L2-018 多项式A除以B
比赛的时候拿不到满分就放弃吧，我已调试到疾苦
想拿满分还是不要看我的这题的一坨解QWQ
#include <bits/stdc++.h>
using namespace std;
#define int long longconst double eps = 1e-6;void ylh_() {int n1;cin >> n1;vector<int> a1;vector<double> a2;for (int i = 1; i <= n1; ++i) {int x;double y;cin >> x >> y;a1.push_back(x);a2.push_back(y);}int n2;cin >> n2;vector<int> b1;vector<double> b2;for (int i = 1; i <= n2; ++i) {int x;double y;cin >> x >> y;b1.push_back(x);b2.push_back(y);}if (n1 == 0) {cout << "0 0 0.0\n 0 0 0.0";return;}// 被除式数组，下标为指数vector<double> a3(10010, 0.0);int max_a = 0;for (int i = 0; i < a1.size(); ++i) {a3[a1[i]] = a2[i];if (a1[i] > max_a) max_a = a1[i];}// 除式数组vector<double> b3(10010, 0.0);int max_b = 0;for (int i = 0; i < b1.size(); ++i) {b3[b1[i]] = b2[i];if (b1[i] > max_b) max_b = b1[i];}// 商数组vector<double> c3(10010, 0.0);int max_c = max_a - max_b;while (max_a >= max_b) {if (abs(a3[max_a]) < eps) {max_a--;continue;}double q = a3[max_a] / b3[max_b];c3[max_a - max_b] = q;// 更新被除式for (int i = max_a, j = max_b; i >= 0 && j >= 0; i--, j--) {a3[i] -= b3[j] * q;}while (max_a >= 0 && abs(a3[max_a]) < eps) {max_a--;}}auto print = [&](vector<double>& arr, int max_exp) {int cnt = 0;for (int i = 0; i <= max_exp; ++i) {if (abs(arr[i]) + 0.05 >= 0.1) cnt++;}cout << cnt;if (cnt == 0) {cout << " 0 0.0";} else {for (int i = max_exp; i >= 0; --i) {if (abs(arr[i]) + 0.05 >= 0.1) {cout << ' ' << i << fixed << setprecision(1) << ' ' << arr[i];}}}};print(c3, max_c);cout << '\n';print(a3, max_a);
}int32_t main() {ios::sync_with_stdio(0);cin.tie(0);int T = 1;// cin >> T;while (T--) {ylh_();}return 0;
}
L2-019 悄悄关注
#include <bits/stdc++.h>
using namespace std;
#define int long long
using PSI = pair<string, int>;void ylh_() {int n;cin >> n;set<string> st;for (int i = 1; i <= n; ++i) {string s;cin >> s;st.insert(s);}int m;cin >> m;vector<PSI> a(m + 1);int sum = 0;for (int i = 1; i <= m; ++i) {cin >> a[i].first >> a[i].second;sum += a[i].second;}sort(a.begin() + 1, a.end(), [&](PSI x, PSI y) {return x.first < y.first;});vector<string> ans;for (int i = 1; i <= m; ++i) {if (a[i].second * m > sum) {if (!st.count(a[i].first)) {ans.push_back(a[i].first);}}}for (int i = 0; i < ans.size(); ++i) {cout << ans[i];if (i != ans.size() - 1) {cout << '\n';} else {return;}}cout << "Bing Mei You";
}int32_t main() {ios::sync_with_stdio(0);cin.tie(0);int T = 1;// cin >> T;while (T--) {ylh_();}
}
L2-020 功夫传人
#include <bits/stdc++.h>
using namespace std;
#define int long longvoid ylh_() {int n;double z, r;cin >> n >> z >> r;r /= 100;vector<vector<int>> g(n);vector<int> ddz(n + 1); // 得道者的倍数for (int i = 0; i < n; ++i) {int m;cin >> m;if (m == 0) {int k;cin >> k;ddz[i] = k;}for (int j = 1; j <= m; ++j) {int k;cin >> k;g[i].push_back(k);}}double ans = 0;auto dfs = [&](auto&& dfs, double cur, int u) {// cout << u << ' ' << cur << endl;if (ddz[u] > 0) {ans += cur * ddz[u];return;}double ncur = cur * (1.0 - r);for (auto v : g[u]) {dfs(dfs, ncur, v);}};dfs(dfs, z, 0);int ans_int = ans;cout << ans_int;
}int32_t main() {ios::sync_with_stdio(0);cin.tie(0);int T = 1;// cin >> T;while (T--) {ylh_();}
}