目录

noip模拟59

T1 柱状图

你可以发现一个规律:考虑以一个最高的柱为分界的左部分和右部分。对于左部分而言,如果$h[mxpos] - mxpos == h[i] - i$ 就可以说明这个点在以$h[mxpos]$为最高高度的屋顶的左半部分。右边把减号换成加号就行。这样就可以判断点在一条斜率为1或-1的直线上下的分布情况。这样就可以二分最高高度了。

对于每个$h[i] - i$和$h[i] + i$都开一棵权值动态开点线段树就行了

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#include <bits/stdc++.h>
#define For(i, l, r) for (int i = (l), i##end = (r); i <= i##end; ++i)
#define Fordown(i, r, l) for (int i = (r), i##end = (l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
const int maxn = 100100;

#define int int64_t

template <typename LDXIN>
LDXIN read() {
  LDXIN s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}

class tree {
 private:
  int num;
  int ls[maxn*30], rs[maxn*30];

 public:
  int rt;
  int64_t sum[maxn*30], sz[maxn*30];
  void insert(int &x, int l, int r, int pos, int val) {
    if (!x) {
      x = ++num;
    }
    if (l == r) {
      sum[x] += 1ll * val * pos;
      sz[x] += val;
      return;
    }
    int mid = (l + r) / 2;
    if (mid <= 0 && l < 0) mid--;
    if (pos <= mid) {
      insert(ls[x], l, mid, pos, val);
    } else {
      insert(rs[x], mid + 1, r, pos, val);
    }
    sum[x] = sum[ls[x]] + sum[rs[x]];
    sz[x] = sz[ls[x]] + sz[rs[x]];
  }
  int query_sz(int x, int l, int r, int ql, int qr) {
    if (!x) return 0;
    if (ql <= l && r <= qr) {
      return sz[x];
    }
    int mid = (l + r) / 2;
    if (mid <= 0 && l < 0) mid--;
    int ans = 0;
    if (ql <= mid) {
      ans += query_sz(ls[x], l, mid, ql, qr);
    }
    if (qr > mid) {
      ans += query_sz(rs[x], mid + 1, r, ql, qr);
    }
    return ans;
  }
  int64_t query_sum(int x, int l, int r, int ql, int qr) {
    if (!x) return 0;
    if (ql <= l && r <= qr) {
      return sum[x];
    }
    int mid = (l + r) / 2;
    if (mid <= 0 && l < 0) mid--;
    int64_t ans = 0;
    if (ql <= mid) {
      ans += query_sum(ls[x], l, mid, ql, qr);
    }
    if (qr > mid) {
      ans += query_sum(rs[x], mid + 1, r, ql, qr);
    }
    return ans;
  }
} ltr, rtr;


int n;
int h[maxn];

int L = -1e5, R = 1e9 + 1e5;

int64_t getAns(int hh, int pos) {
  int64_t ans = 0;
  ans += 1ll * ltr.query_sz(ltr.rt, L, R, L, hh - pos - 1) * (hh - pos) -
         ltr.query_sum(ltr.rt, L, R, L, hh - pos - 1);
  ans += 1ll * rtr.query_sz(rtr.rt, L, R, L, hh + pos - 1) * (hh + pos) -
         rtr.query_sum(rtr.rt, L, R, L, hh + pos - 1);
  ans += ltr.query_sum(ltr.rt, L, R, hh - pos + 1, R) -
         1ll * ltr.query_sz(ltr.rt, L, R, hh - pos + 1, R) * (hh - pos);
  ans += rtr.query_sum(rtr.rt, L, R, hh + pos + 1, R) -
         1ll * rtr.query_sz(rtr.rt, L, R, hh + pos + 1, R) * (hh + pos);
  return ans;
}

int32_t main() {
  freopen("c.in", "r", stdin);
  freopen("c.out", "w", stdout);
  cin >> n;
  int mx = 0, mxid;
  For(i, 1, n) {
    cin >> h[i];
    if (h[i] > mx) {
      mx = h[i];
      mxid = i;
    }
    rtr.insert(rtr.rt, L, R, h[i] + i, 1);
  }
  int64_t ans = 0x3f3f3f3f3f3f3f3f;
  For(i, 1, n) {
    rtr.insert(rtr.rt, L, R, h[i] + i, -1);
    ltr.insert(ltr.rt, L, R, h[i] - i, 1);
    int l = max(i, n - i + 1), r = max(l, mx + abs(i - mxid));
    while (r - l > 1) {
      int mid = (l + r) / 2;
      int lwr = ltr.query_sz(ltr.rt, L, R, L, mid - i - 1) +
                rtr.query_sz(rtr.rt, L, R, L, mid + i - 1);
      int upr = n - lwr;
      if (upr >= lwr) {
        l = mid;
      } else {
        r = mid;
      }
    }
    ans = min(ans, min(getAns(l, i), getAns(r, i)));
  }
  cout << ans << endl;
  return 0;
}

T2 应急棍

100分的高精小数不会打……

这是一道找规律题,除了前四个点以外,剩下的都是先放在中心,再放在边上。而随着边长的增加,中心的点的个数为$4^{n - 1}$,在边上的点个数规律是这样的:$1,2,1 \quad 2,3,2,3,2 \quad 4,5,4,5,4,5,4,5,4$所以第一列的点数为$2^{n - 1}$下一列的点数为$2^{n - 1} + 1$

然后……

码就完事了

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#include <bits/stdc++.h>
#define For(i, l, r) for (int i = (l), i##end = (r); i <= i##end; ++i)
#define Fordown(i, r, l) for (int i = (r), i##end = (l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;

#define int int64_t

double l, c;
int32_t main() {
  freopen("a.in", "r", stdin);
  freopen("a.out", "w", stdout);
  int t;
  cin >> c >> t >> l;
  cout.flags(ios::fixed);
  cout.precision(c);
  For(i, 1, t) {
    int id;
    cin >> id;
    if (id == 1) {
      cout << 0 << ' ' << 0 << endl;
      continue;
    }
    if (id == 2) {
      cout << l << ' ' << l << endl;
      continue;
    }
    if (id == 3) {
      cout << 0 << ' ' << l << endl;
      continue;
    }
    if (id == 4) {
      cout << l << ' ' << 0 << endl;
      continue;
    }
    int lastno = log2(floor(sqrt(id - 1)) - 1);
    int remain;
    if (lastno == 0) {
      remain = id - 4;
    } else {
      remain = id - (pow(2, lastno) + 1) * (pow(2, lastno) + 1);
    }
    int sid = pow(4, lastno);
    int sLine = pow(2, lastno);
    int oLine = sLine + 1;
    // cout << sLine << endl;
    double dlt = l / sLine / 2;
    if (remain <= sid) {
      int cnt = 0;
      while (remain > sLine) {
        remain -= sLine;
        cnt++;
      }
      double x = dlt + dlt * cnt * 2;
      double y = dlt + (remain - 1) * dlt * 2;
      cout << x << ' ' << y << endl;
      continue;
    }

    remain -= sid;
    double x, y;
    int cnt = 0;
    while (remain) {
      if (cnt % 2) {
        if (remain <= oLine) break;

        remain -= oLine;
        cnt++;
        if (remain < sLine) break;
      } else {
        if (remain <= sLine) break;
        remain -= sLine;
        cnt++;
        if (remain < oLine) break;
      }
    }
    x = dlt * cnt;
    if (cnt % 2)
      y = (remain - 1) * dlt * 2;
    else
      y = remain * dlt * 2 - dlt;
    cout << x << ' ' << y << endl;
  }
  return 0;
}