Luogu 3793 由乃救爷爷

\(\verb|Luogu 3793 由乃救爷爷|\)

rmq，数据随机

\(n,\ m\leq 2\times10^7\)

lxl ST表

分块，大小设为 \(x\)

预处理每个块两端到块内每个点的前缀 \(\max\) 和后缀 \(\max\)

预处理块间ST表

数据随机

就成了期望 \(O(n)\) 的rmq

重点是没人卡你，卡还不一定卡的住，还要冒着被暴力AC的风险

然后就愉悦地~

#include <bits/stdc++.h>
using namespace std;

#define get(x) (((x) + 63) >> 6)
typedef unsigned uint;
const int maxn = 2e7 + 10;
int n, m, K, tot, a[maxn], lef[maxn], rig[maxn], lg[maxn >> 6], val[20][maxn >> 6];

namespace GenHelper {
  uint z1, z2, z3, z4, b;
  inline uint rnd() {
    b = ((z1 << 6) ^ z1) >> 13;
    z1 = ((z1 & 4294967294u) << 18) ^ b;
    b = ((z2 << 2) ^ z2) >> 27;
    z2 = ((z2 & 4294967288u) << 2) ^ b;
    b = ((z3 << 13) ^ z3) >> 21;
    z3 = ((z3 & 4294967280u) << 7) ^ b;
    b = ((z4 << 3) ^ z4) >> 12;
    z4 = ((z4 & 4294967168u) << 13) ^ b;
    return (z1 ^ z2 ^ z3 ^ z4);
  }
}

inline void srand(uint x) {
  using namespace GenHelper;
  z1 = x;
  z2 = (~x) ^ 0x233333333u;
  z3 = x ^ 0x1234598766u;
  z4 = (~x) + 51;
}

inline int read() {
  using namespace GenHelper;
  int a = rnd() & 32767;
  int b = rnd() & 32767;
  return a << 15 | b;
}

inline int query(int l, int r) {
  if (l > r) return 0;
  int tmp = lg[r - l + 1];
  return max(val[tmp][l], val[tmp][r - (1 << tmp) + 1]);
}

int main() {
  scanf("%d %d %d", &n, &m, &K);
  srand(K);
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    val[0][get(i)] = max(a[i], val[0][get(i)]);
  }
  tot = get(n);
  for (int i = 2; i <= tot; ++i) {
    lg[i] = lg[i >> 1] + 1;
  }
  for (int i = 1, lst = 0; i <= n; ++i) {
    lef[i] = lst = max(a[i], lst);
    if (!(i & 63)) lst = 0;
  }
  for (int i = n, lst = 0; i; --i) {
    if (!(i & 63)) lst = 0;
    rig[i] = lst = max(a[i], lst);
  }
  for (int i = 1; i < 21; ++i) {
    for (int j = 1; j + (1 << i) <= tot; ++j) {
      val[i][j] = max(val[i - 1][j], val[i - 1][j + (1 << (i - 1))]);
    }
  }
  unsigned long long ans = 0;
  for (int q = 1, l, r, L, R, res; q <= m; ++q) {
    l = read() % n + 1;
    r = read() % n + 1;
    if (l > r) l ^= r ^= l ^= r;
    L = get(l), R = get(r), res = 0;
    if (L == R) {
      for (int i = l; i <= r; ++i) {
        res = max(res, a[i]);
      }
    } else {
      res = max(max(rig[l], lef[r]), query(L + 1, R - 1));
    }
    ans += res;
  }
  printf("%llu", ans);
  return 0;
}

原文地址：https://www.cnblogs.com/Juanzhang/p/10463796.html