萌新求助,线段树1那题,本地AC提交RE

P3372 【模板】线段树 1

把没有返回值的函数改成 void 就行了。不返回开了 O2 会 RE。
by xhgua @ 2024-03-23 13:58:30


@[xhgua](/user/410239) 是 Linux 特有的吗( 拜谢 [xhgua](/user/410239)
by CEFqwq @ 2024-03-23 14:01:48


@[爱肝大模拟的tlxjy](/user/482610) 未定义行为,不开 O2 大概率没事。
by xhgua @ 2024-03-23 14:03:15


@[xhgua](/user/410239) 我本地开了 O2 没事,洛谷不开 O2 出事/yun
by CEFqwq @ 2024-03-23 14:04:42


@[爱肝大模拟的tlxjy](/user/482610) 可能版本比较低,我本地开了 O2 会挂
by xhgua @ 2024-03-23 14:08:48


@[xhgua](/user/410239) 你是什么版本…… 我 Dev-C++5.11 ISOC++11 O2
by CEFqwq @ 2024-03-23 14:09:47


@[爱肝大模拟的tlxjy](/user/482610) 很难绷,我版本比你还低... 那有点奇怪...
by xhgua @ 2024-03-23 14:11:53


@[xhgua](/user/410239) 好像我还有哪里 UB 了/yun/yun/yun ```cpp #include<bits/stdc++.h> #define int long long using namespace std; #define mxn 200005 struct node{ int l, r, vl, lzy; }tr[mxn << 2 | 3]; int a[mxn], n, q; int ls(int u){ return u << 1; } int rs(int u){ return u << 1 | 1; } void glz(int u, int k){ tr[u].lzy += k; } void gv(int u, int k){ tr[u].vl += k; } int md(int l, int r){ return (l + r) >> 1; } void pup(int u){ tr[u].vl = tr[ls(u)].vl + tr[rs(u)].vl; } void build(int u, int l, int r){ tr[u].l = l; tr[u].r = r; if (l == r){ tr[u].vl = a[l]; return; } int mid = md(l, r); build(ls(u), l, mid); build(rs(u), mid + 1, r); pup(u); } bool inr(int l, int r, int ql, int qr){ return (ql <= l && qr >= r); } bool otr(int l, int r, int ql, int qr){ return (ql > r || qr < l); } void pdn(int u, int l, int r){ int mid = md(l, r); glz(ls(u), tr[u].lzy); gv(ls(u), tr[u].lzy * (mid - l + 1)); glz(rs(u), tr[u].lzy); gv(rs(u), tr[u].lzy * (r - (mid + 1) + 1)); tr[u].lzy = 0; } void upd(int u, int ql, int qr, int x){ if (inr(tr[u].l, tr[u].r, ql, qr)){ glz(u, x); gv(u, x * (tr[u].r - tr[u].l + 1)); return; } if (otr(tr[u].l, tr[u].r, ql, qr)) return; pdn(u, ql, qr); int mid = md(tr[u].l, tr[u].r); upd(ls(u), ql, qr, x); upd(rs(u), ql, qr, x); } int qry(int u, int ql, int qr){ int res = 0; if (inr(tr[u].l, tr[u].r, ql, qr)) return tr[u].vl; if (otr(tr[u].l, tr[u].r, ql, qr)) return 0; pdn(u, ql, qr); int mid = md(tr[u].l, tr[u].r); res += (qry(ls(u), ql, qr) + qry(rs(u), ql, qr)); return res; } signed main(){ ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); cin >> n >> q; for (int i = 1; i <= n; i++) cin >> a[i]; build(1, 1, n); while (q--){ int op; cin >> op; if (op == 1){ int l, r, x; cin >> l >> r >> x; upd(1, l, r, x); }else{ int l, r; cin >> l >> r; cout << qry(1, l, r) << "\n"; } // for (int i = 1; i <= n; i++) // cout << qry(1, i, i) << " "; // cout << "\n"; } } ```
by CEFqwq @ 2024-03-23 14:23:23


u7y
by 094913a @ 2024-03-23 14:56:18


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