给定一个数的序列S,以及一个区间[L, R], 求序列中介于该区间的数的个数,即序列中大于等于L且小于等于R的数的个数。C++
时间: 2023-12-02 20:43:12 浏览: 62
以下是C++代码实现:
```cpp
#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int w[N];
struct Node
{
int l, r;
int cnt;
}tr[N * 4];
void pushup(int u)
{
tr[u].cnt = tr[u << 1].cnt + tr[u << 1 | 1].cnt;
}
void build(int u, int l, int r)
{
if (l == r) tr] = {l, r, 0};
else
{
tr[u] = {l, r};
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}
void modify(int u, int x)
{
if (tr[u].l == x && tr[u].r == x) tr[u].cnt++;
else
{
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x);
else modify(u << 1 | 1, x);
pushup(u);
}
}
int query(int u, int l, int r)
{
if (tr[u].l >= l && tr[u].r <= r) return tr[u].cnt;
else
{
int mid = tr[u].l + tr[u].r >> 1;
int res = 0;
if (l <= mid) res += query(u << 1, l, r);
if (r > mid) res += query(u << 1 | 1, l, r);
return res;
}
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> w[i];
build(1, 1, n);
while (m--)
{
int l, r;
cin >> l >> r;
cout << query(1, l, r) << endl;
}
return 0;
}
```