Blog of RuSun

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

P2801 教主的魔法

P2801 教主的魔法

注意到 $n$ 很大,$q$ 很小。

考虑分块。块内排序。可以在 $O(\log {\sqrt n})$ 的时间查询一块的答案,$O(1)$ 的时间修改一块。对于两端的修改,重构整块,复杂度 $O(\sqrt n \log {\sqrt n})$ 。

查看代码
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#include <cstdio>
#include <cmath>
#include <algorithm>
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 << 3) + (x << 1) + c - '0';
(flag) && (x = ~x + 1);
}
const int N = 1e6 + 10;
int n, m, len, w[N], v[N], d[N];
void build(int x)
{
for (int i = x * len; i < min(n + 1, x * len + len); i++)
v[i] = w[i];
sort(v + x * len, v + min(n + 1, x * len + len));
}
void modify(int l, int r, int k)
{
if (l / len == r / len)
{
for (int i = l; i <= r; i++)
w[i] += k;
build(l / len);
return;
}
for (int i = l; i < l / len * len + len; i++)
w[i] += k;
build(l / len);
for (int i = r / len * len; i <= r; i++)
w[i] += k;
build(r / len);
for (int i = l / len + 1; i < r / len; i++)
d[i] += k;
}
int query(int l, int r, int k)
{
int res = 0;
if (l / len == r / len)
{
for (int i = l; i <= r; i++)
if (w[i] + d[l / len] >= k)
res++;
return res;
}
for (int i = l; i < l / len * len + len; i++)
res += w[i] + d[l / len] >= k;
for (int i = r / len * len; i <= r; i++)
res += w[i] + d[r / len] >= k;
for (int i = l / len + 1; i < r / len; i++)
res += i * len + len - (lower_bound(v + i * len, v + i * len + len, k - d[i]) - v);
return res;
}
int main()
{
read(n), read(m);
len = sqrt(n);
for (int i = 1; i <= n; i++)
read(w[i]);
for (int i = 0; i <= n / len; i++)
build(i);
for (char op; m; m--)
{
int l, r, k;
op = getchar();
while (op != 'A' && op != 'M')
op = getchar();
read(l), read(r), read(k);
if (op == 'M')
modify(l, r, k);
else if (op == 'A')
printf("%d\n", query(l, r, k));
}
return 0;
}