|
|
#include <bits/stdc++.h>
|
|
|
using namespace std;
|
|
|
int const N = 50010;
|
|
|
int n, m;
|
|
|
|
|
|
// 线段树+单点修改
|
|
|
#define mid ((l + r) >> 1)
|
|
|
#define ls (u << 1)
|
|
|
#define rs (u << 1 | 1)
|
|
|
struct Node {
|
|
|
int l, r;
|
|
|
int sum;
|
|
|
} tr[2][N << 2];
|
|
|
|
|
|
void pushup(int w, int u) {
|
|
|
tr[w][u].sum = tr[w][ls].sum + tr[w][rs].sum;
|
|
|
}
|
|
|
void build(int w, int u, int l, int r) {
|
|
|
tr[w][u].l = l, tr[w][u].r = r;
|
|
|
if (l == r) return;
|
|
|
build(w, ls, l, mid);
|
|
|
build(w, rs, mid + 1, r);
|
|
|
}
|
|
|
|
|
|
void change(int w, int u, int x, int v) {
|
|
|
int l = tr[w][u].l, r = tr[w][u].r;
|
|
|
if (l == r) {
|
|
|
tr[w][u].sum += v;
|
|
|
return;
|
|
|
}
|
|
|
if (x <= mid)
|
|
|
change(w, ls, x, v);
|
|
|
else
|
|
|
change(w, rs, x, v);
|
|
|
pushup(w, u);
|
|
|
}
|
|
|
|
|
|
int query(int w, int u, int L, int R) {
|
|
|
int l = tr[w][u].l, r = tr[w][u].r;
|
|
|
if (l >= L && r <= R) return tr[w][u].sum;
|
|
|
if (l > R || r < L) return 0;
|
|
|
return query(w, ls, L, R) + query(w, rs, L, R);
|
|
|
}
|
|
|
|
|
|
int main() {
|
|
|
#ifndef ONLINE_JUDGE
|
|
|
freopen("LOJ10115.in", "r", stdin);
|
|
|
#endif
|
|
|
// 加快读入
|
|
|
ios::sync_with_stdio(false), cin.tie(0);
|
|
|
int op, l, r;
|
|
|
cin >> n >> m; // 第一行 n,m 表示道路总长为 n，共有 m 个操作；
|
|
|
// n下面没有使用过。为什么呢？其实是n的上限N有用！我们就没有用到n,代码模板中也去掉了n的
|
|
|
build(0, 1, 1, n);
|
|
|
build(1, 1, 1, n);
|
|
|
|
|
|
while (m--) {
|
|
|
cin >> op >> l >> r;
|
|
|
if (op == 1)
|
|
|
change(0, 1, l, 1), change(1, 1, r, 1);
|
|
|
else
|
|
|
printf("%d\n", query(0, 1, 1, r) - query(1, 1, 1, l - 1));
|
|
|
}
|
|
|
return 0;
|
|
|
} |
