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Commit b7c120f

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灵茶の试炼 * 41 (无UT)
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‎algo-hub-cpp/CMakeLists.txt‎

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@@ -14,3 +14,6 @@ add_executable(P3594 [POI2015] WIL "luogu/P3594 [POI2015] WIL.cpp")
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add_executable(P3901 数列找不同 "luogu/P3901 数列找不同.cpp")
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add_executable(P4513 小白逛公园 "luogu/P4513 小白逛公园.cpp")
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add_executable(P5490 【模板】扫描线 & 矩形面积并 "luogu/P5490 【模板】扫描线 & 矩形面积并.cpp")
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add_executable(P3572 [POI 2014] PTA-Little Bird "luogu/P3572 [POI 2014] PTA-Little Bird.cpp")
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add_executable(P3694 邦邦的大合唱站队 "luogu/P3694 邦邦的大合唱站队.cpp")
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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struct pair1 {
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int f, i;
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};
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void solve() {
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int n;
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cin >> n;
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vector<int> a(n);
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for (int i = 0; i < n; ++i) {
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cin >> a[i];
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}
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int t;
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cin >> t;
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while (t-- > 0) {
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int k;
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cin >> k;
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deque<pair1> dq; // f, i
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dq.push_back({0, 0});
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for (int i = 1; i < n; i++) {
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if (dq.front().i < i - k) {
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dq.pop_front();
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}
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int f = dq.front().f;
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if (a[i] >= a[dq.front().i]) {
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f++;
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}
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while (!dq.empty() && f < dq.back().f || f == dq.back().f && a[i] >= a[dq.back().i]) {
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dq.pop_back();
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}
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dq.push_back({f, i});
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}
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cout << dq.back().f << endl;
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}
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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void solve() {
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int n, m;
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cin >> n >> m;
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vector<int> a(n);
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for (int i = 0; i < n; ++i) {
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cin >> a[i];
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}
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vector<vector<int> > sum(n + 1, vector<int>(20));
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for (int i = 1; i <= n; i++) {
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int v = a[i - 1];
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sum[i] = sum[i - 1];
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sum[i][v - 1]++;
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}
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int u = 1 << m;
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vector<int> sz(u);
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for (int i = 0; i < m; i++) {
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int v = sum[n][i];
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int highBit = 1 << i;
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for (int mask = 0; mask < highBit; mask++) {
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int s = sz[mask];
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sz[highBit | mask] = s + v;
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}
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}
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vector<int> f(u);
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for (int s = 0; s < u; s++) {
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int fs = f[s];
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// for cus, lb := u-1^s, 0; cus > 0; cus ^= lb {
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for (int cus = u - 1 ^ s, lb = 0; cus > 0; cus ^= lb) {
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lb = cus & -cus;
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int ns = s | lb;
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int p = __builtin_ctz(lb);
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f[ns] = max(f[ns], fs + sum[sz[ns]][p] - sum[sz[s]][p]);
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}
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}
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int ans = n - f[u - 1];
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cout << ans << endl;
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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*/
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package c367;
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import java.util.HashMap;
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import java.util.Map;
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import java.util.Scanner;
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public class Abc367_d {
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static int n, m;
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static int[] a;
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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n = scanner.nextInt();
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m = scanner.nextInt();
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a = new int[n];
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for (int i = 0; i < n; i++) {
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a[i] = scanner.nextInt();
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}
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System.out.println(solve());
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}
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private static String solve() {
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long sl = 0, sr = 0, ans = 0;
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Map<Long, Integer> cnt = new HashMap<>();
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for (int i = 0; i < n * 2 - 1; i++) {
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sr += a[i % n];
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if (i >= n - 1) {
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ans += cnt.getOrDefault(sr % m, 0);
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sl += a[i - n + 1];
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cnt.merge(sl % m, -1, Integer::sum);
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}
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cnt.merge(sr % m, 1, Integer::sum);
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}
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return String.valueOf(ans);
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}
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}
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/*
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D - Pedometer
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https://atcoder.jp/contests/abc367/tasks/abc367_d
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灵茶の试炼 2025年05月13日
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题目大意:
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输入 n(2≤n≤2e5) m(1≤m≤1e6) 和长为 n 的数组 a(1≤a[i]≤1e9)。下标从 1 开始。
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一个环上有 n 个位置,顺时针编号从 1 到 n。
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从 i 顺时针移动到 i+1,需要走 a[i] 步。
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特别地,从 n 顺时针移动到 1,需要走 a[n] 步。
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输出有多少对位置 (s,t),满足 s≠t,且从 s 顺时针移动到 t 的最小步数是 m 的倍数。
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断环为链,把 a 变成一个长为 2n-1 的数组 b。
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问题变成求有 b 中有多少个长度 <= n-1 的非空子数组,其元素和是 m 的倍数。
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如果你对此没有任何思路,可以做做我的 数据结构题单 中的 §1.2 前缀和与哈希表
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计算 b 的前缀和数组。用哈希表统计前缀和的出现次数。
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(下标从 0 开始)枚举在 [n-1,2n-2] 中的 t,计算有多少个符合要求的 s。
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如何保证长度 <= n-1?可以在枚举 t 的过程中,从哈希表中删除(出现次数减一)距离 t 太远的 s 的前缀和。(类似定长滑窗)
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代码 https://atcoder.jp/contests/abc367/submissions/64678165
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======
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Input 1
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4 3
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2 1 4 3
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Output 1
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4
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Input 2
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2 1000000
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1 1
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Output 2
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0
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Input 3
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9 5
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9 9 8 2 4 4 3 5 3
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Output 3
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11
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*/

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