大冤种。
第一眼拿到题目：这不一眼前缀和吗？
结果没看时限&没分析时间复杂度的我：前缀和肯定会 T 飞，需要优化。
怎么优化呢？这不一眼线段树吗，于是我写了 \(100\) 多行线段树。
这里的 \(1\) 操作是让我们交换 \(a_x\) 和 \(a_{x+1}\) 的值，那么这里的 update 函数我们可以这么写：
void update(ll d, ll pl, ll pr, ll p){if (pl == pr){tree[p] = a[pl];return;}pushdown(p, pl, pr);ll mid = (pl + pr) >> 1;if (d <= mid)if (d + 1 <= mid)//都在左子树update(d, pl, mid, ls(p));else //一左一右update(d, pl, mid, ls(p)), update(d + 1, mid + 1, pr, rs(p));else //都在右子树update(d, mid + 1, pr, rs(p));pushup(p);
}
\(2\) 操作就是个区间和（前缀和）板子，不多讲了，直接上冤种代码。
/*********************************************************** Author : dingziyang888* Website : https://www.luogu.com.cn/problem/* Created Time :* FileName :* Warning!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!* 1.MLE?* 2.array size enough?* 3.long long?* 4.overflow long long?* 5.multiple task cleaned?* 6.freopen?* 7.TLE?* *******************************************************/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <climits>
#define DEBUG
#define Ofile(s) freopen(s".in", "r", stdin), freopen (s".out", "w", stdout)
#define Cfile(s) fclose(stdin), fclose(stdout)
#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;using ll = long long;
using ull = unsigned long long;
using lb = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;constexpr int mod = 998244353;
constexpr int maxk = 200005;ll n, q, L, R, x, opt;ll a[maxk], tree[maxk << 2], tag[maxk << 2];ll ls(ll p){return p << 1;
}ll rs(ll p){return p << 1 | 1;
}void pushup(int p){tree[p] = tree[ls(p)] + tree[rs(p)];
}void build(ll p, ll pl, ll pr){tag[p] = 0;if (pl == pr){tree[p] = a[pl];return;}ll mid = (pl + pr) >> 1;build(ls(p), pl, mid);build(rs(p), mid + 1, pr);pushup(p);
}void addtag(ll p, ll pl, ll pr, ll d){tag[p] += d;tree[p] += d * (pr - pl + 1);
}void pushdown(ll p, ll pl, ll pr){if (tag[p]){ll mid = (pl + pr) >> 1;addtag(ls(p), pl, mid, tag[p]);addtag(rs(p), mid + 1, pr, tag[p]);tag[p] = 0;}
}ll query(ll L, ll R, ll p, ll pl, ll pr){if (L <= pl && pr <= R)return tree[p];pushdown(p, pl, pr);ll mid = (pl + pr) >> 1, res = 0;if (L <= mid)res += query(L, R, ls(p), pl, mid);if (mid < R)res += query(L, R, rs(p), mid + 1, pr);return res;
}void update(ll d, ll pl, ll pr, ll p){if (pl == pr){tree[p] = a[pl];return;}pushdown(p, pl, pr);ll mid = (pl + pr) >> 1;if (d <= mid)if (d + 1 <= mid)update(d, pl, mid, ls(p));else update(d, pl, mid, ls(p)), update(d + 1, mid + 1, pr, rs(p));else update(d, mid + 1, pr, rs(p));pushup(p);
}int main() {fast;cin >> n >> q;for (int i = 1; i <= n; i++)cin >> a[i];build(1, 1, n);while (q--){cin >> opt;if (opt == 1){cin >> x;swap(a[x], a[x + 1]);update(x, 1, n, 1);update(x + 1, 1, n, 1);}else {cin >> L >> R;cout << query(L, R, 1, 1, n) << endl;}}return 0;
}