Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF840D Destiny

LuoGu: CF840D Destiny

CF: D. Destiny

首先有弱化版,$k = 2$ ,这样相当于区间出现超过一半的数,那么主席树上可以直接通过当前左子树判断向哪棵子树走,但是 $k$ 更大的时候就不行了。注意到满足的条件的数最多有 $k - 1$ 个,所以最多只有 $O(k \log n)$ 的节点是满足条件的,如果满足条件的数更少,那么满足条件的节点会更少。所以遍历这些节点可以保证复杂度。

查看代码
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#include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
const int N = 3e5 + 10, M = 6e6 + 10;
struct Node
{
int l, r, s;
} tr[M];
int n, m, idx, rt[N];
void modify (int &x, int t, int l = 1, int r = n)
{
tr[++idx] = tr[x];
tr[x = idx].s++;
if (l == r)
return;
int mid = l + r >> 1;
t <= mid ? modify(tr[x].l, t, l, mid) : modify(tr[x].r, t, mid + 1, r);
}
int query (int x, int y, int k, int l = 1, int r = n)
{
if (tr[x].s - tr[y].s <= k)
return -1;
if (l == r)
return l;
int mid = l + r >> 1;
int t = query(tr[x].l, tr[y].l, k, l, mid);
return ~t ? t : query(tr[x].r, tr[y].r, k, mid + 1, r);
}
int main ()
{
read(n), read(m);
for (int i = 1, a; i <= n; i++)
read(a), modify(rt[i] = rt[i - 1], a);
for (int l, r, k; m; m--)
{
read(l), read(r), read(k);
write(query(rt[r], rt[l - 1], (r - l + 1) / k)), puts("");
}
return 0;
}