113 lines
3.4 KiB
C++
113 lines
3.4 KiB
C++
#include <iostream>
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using namespace std;
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typedef long long LL;
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#define ls u << 1
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#define rs u << 1 | 1
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const int N = 100010;
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//快读
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int read() {
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int x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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int n, q;
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struct Node {
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int l, r;
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LL sum, tag; //区间总和,修改的数值(懒标记)
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} tr[N << 2];
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//更新统计信息
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void pushup(int u) {
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tr[u].sum = tr[ls].sum + tr[rs].sum;
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}
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//父节点向子孙节点传递信息
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void pushdown(int u) {
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if (tr[u].tag) {
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//更新儿子的tag信息,是为了更下一层的数据传递
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// tag传递到子段,节段的sum和需要按 区间长度*root.tag 进行增加
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tr[ls].tag += tr[u].tag;
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tr[ls].sum += LL(tr[ls].r - tr[ls].l + 1) * tr[u].tag;
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tr[rs].tag += tr[u].tag;
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tr[rs].sum += LL(tr[rs].r - tr[rs].l + 1) * tr[u].tag;
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//别忘了清除懒标记
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tr[u].tag = 0;
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}
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}
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//构建
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void build(int u, int l, int r) {
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tr[u].l = l, tr[u].r = r; //标记范围
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if (l == r) { //叶子
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tr[u].sum = read(); //区间内只有一个元素l(r),区间和为read(),不需要记录向下的传递tag
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return;
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}
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int mid = (l + r) >> 1;
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build(ls, l, mid), build(rs, mid + 1, r); //左右儿子构建
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pushup(u); //通过左右儿子构建后,向祖先节点反馈统计信息变化
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}
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//以u为根,在区间[l,r]之间全都增加d
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void add(int u, int l, int r, int d) {
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if (tr[u].l >= l && tr[u].r <= r) {
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tr[u].sum += LL(tr[u].r - tr[u].l + 1) * d; //总和增加
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tr[u].tag += d; //懒标记+d
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return;
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}
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pushdown(u); // 如果自己身上有旧的tag数值,在递归前需要将原tag值pushdown到子孙节点去
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int mid = tr[u].l + tr[u].r >> 1;
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if (l <= mid) add(ls, l, r, d); //与左区间有交集
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if (r > mid) add(rs, l, r, d); //与右区间有交集
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pushup(u); //将结果的变更更新到祖先节点
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}
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//关键的查询操作
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LL query(int u, int l, int r) {
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if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
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/*懒标记的思想:在更新时只标识不修改,有查询时现用现改的思想。这么做的目的当然是怕有多组修改,
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就需要修改多次,而且路径太长不划算。如果只标识不修改,直接返回,当然快了。比如修改了3次后查询一次,
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也只是真正修改了一次,降低了修改次数。
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*/
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pushdown(u);
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int mid = tr[u].l + tr[u].r >> 1;
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LL res = 0;
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if (l <= mid) res = query(ls, l, r);
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if (r > mid) res += query(rs, l, r);
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//左查+ 右查 = 总和
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return res;
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}
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int main() {
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//文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("P3372.in", "r", stdin);
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#endif
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n = read(), q = read();
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//构建线段树
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build(1, 1, n);
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while (q--) {
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int a, b, c, d;
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a = read(), b = read(), c = read();
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if (b > c) swap(b, c);
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if (a == 1)
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d = read(), add(1, b, c, d);
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else
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printf("%lld\n", query(1, b, c));
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}
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return 0;
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} |