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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#520849#7605. Yet Another Mex ProblemliuziaoWA 5ms45256kbC++144.1kb2024-08-15 16:31:472024-08-15 16:31:48

Judging History

你现在查看的是最新测评结果

  • [2024-08-15 16:31:48]
  • 评测
  • 测评结果:WA
  • 用时:5ms
  • 内存:45256kb
  • [2024-08-15 16:31:47]
  • 提交

answer

#include <bits/stdc++.h>

#define int int64_t

using pii = std::pair<int, int>;

const int kMaxN = 2e5 + 5;

int n, k;
int a[kMaxN], sum[kMaxN], f[kMaxN];

pii sub(pii a, pii b) { return {a.first - b.first, a.second - b.second}; }
int mul(pii a, pii b) { return a.first * b.second - a.second * b.first; }
int func(pii a, int k) { return a.first * k + a.second; }

struct SGT1 {
  int mi[kMaxN * 4];

  void pushup(int x) { mi[x] = std::min(mi[x << 1], mi[x << 1 | 1]); }

  void update(int x, int l, int r, int ql, int v) {
    if (l == r) return void(mi[x] = v);
    int mid = (l + r) >> 1;
    if (ql <= mid)
      update(x << 1, l, mid, ql, v);
    else
      update(x << 1 | 1, mid + 1, r, ql, v);
    pushup(x);
  }

  int getmex(int x, int l, int r, int ql) {
    assert(mi[x] < ql);
    if (l == r) return l;
    int mid = (l + r) >> 1;
    if (mi[x << 1] < ql)
      return getmex(x << 1, l, mid, ql);
    else
      return getmex(x << 1 | 1, mid + 1, r, ql);
  }

  int query(int x, int l, int r, int ql, int qr) {
    if (l > qr || r < ql)
      return 1e9;
    else if (l >= ql && r <= qr)
      return mi[x];
    int mid = (l + r) >> 1;
    return std::min(query(x << 1, l, mid, ql, qr),
                    query(x << 1 | 1, mid + 1, r, ql, qr));
  }
} sgt1;

struct SGT2 {  // f[l] - mex * sum[l]
  // (sum[l], f[l]) 上凸壳

  std::vector<pii> v[kMaxN * 4];

  void ins(int x, pii p) {
    // if (!v[x].empty() && p <= v[x].back()) return;
    // for (;
    //      v[x].size() >= 2 &&
    //      mul(sub(p, v[x].back()), sub(v[x].back(), v[x][v[x].size() - 2])) <= 0;
    //      v[x].pop_back()) {
    // }
    v[x].emplace_back(p);
  }

  int getmax(int x, int k) {
    if (!v[x].size()) return -1e18;
    // int L = 0, R = v[x].size(), res = 0;
    // while (L + 1 < R) {
    //   int mid = (L + R) >> 1;
    //   if (func(v[x][mid], k) >= func(v[x][mid - 1], k))
    //     L = res = mid;
    //   else
    //     R = mid;
    // }
    // return func(v[x][res], k);
    int ret = -1e18;
    for (auto p : v[x]) ret = std::max(ret, func(p, k));
    return ret;
  }

  void update(int x, int l, int r, int ql, pii p) {
    ins(x, p);
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (ql <= mid)
      update(x << 1, l, mid, ql, p);
    else
      update(x << 1 | 1, mid + 1, r, ql, p);
  }

  int query(int x, int l, int r, int ql, int qr, int k) {
    if (l > qr || r < ql)
      return -1e18;
    else if (l >= ql && r <= qr)
      return getmax(x, k);
    int mid = (l + r) >> 1;
    return std::max(query(x << 1, l, mid, ql, qr, k),
                    query(x << 1 | 1, mid + 1, r, ql, qr, k));
  }
} sgt2, sgt3;

void dickdreamer() {
  std::cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i];
    sum[i] = sum[i - 1] + a[i];
  }
  sgt2.update(1, 0, n, 0, {0, 0}), sgt3.update(1, 0, n, 0, {0, 0});
  for (int i = 1; i <= n; ++i) {
    int l = sgt1.query(1, 0, n, 0, a[i]) + 1,
        r = (a[i] ? sgt1.query(1, 0, n, 0, a[i] - 1) : i);
    sgt1.update(1, 0, n, a[i], i);
    // std::cerr << i << " : ---------\n";
    for (; l <= r;) {
      int mex = sgt1.getmex(1, 0, n, r), t = sgt1.query(1, 0, n, 0, mex);
      int val = sgt2.query(1, 0, n, t, r - 1, -mex);
      sgt3.update(1, 0, n, t, {mex, val});
      r = t;
    }
    l = std::max<int>(i - k + 1, 1);
    int mex = sgt1.getmex(1, 0, n, l);
    r = (mex ? sgt1.query(1, 0, n, 0, mex - 1) : i);
    assert(l <= r);
    f[i] = std::max(sgt3.query(1, 0, n, l - 1, i - 1, sum[i]),
                    sgt2.query(1, 0, n, l - 1, r - 1, -mex) + sum[i] * mex);
    sgt2.update(1, 0, n, i, {sum[i], f[i]});
    sgt3.update(1, 0, n, i, {!a[i], f[i]});
    // std::cerr << i << ' ' << f[i] << '\n';
  }
  std::cout << f[n] << '\n';
}

int32_t main() {
#ifdef ORZXKR
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
#endif
  std::ios::sync_with_stdio(0), std::cin.tie(0), std::cout.tie(0);
  int T = 1;
  // std::cin >> T;
  while (T--) dickdreamer();
  // std::cerr << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
  return 0;
}

詳細信息

Test #1:

score: 0
Wrong Answer
time: 5ms
memory: 45256kb

input:

5 3
3 4 0 0 3

output:

24

result:

wrong answer 1st numbers differ - expected: '10', found: '24'