基础&进阶线段树学习笔记（一） | P3372 【模板】线段树 1 题解( 二 ) _生活百科

查询其实和修改很像。（不妨肉眼观查一下）在该结点所管辖的区间完全被覆盖时，直接返回区间和。若未被完全覆盖，则下传懒标记，分左右儿子考虑，统计总答案并返回值。是不是 so easy？
于是你愉快地切了这题。

完整代码，点击查看

#include <bits/stdc++.h>using namespace std;#define ll long long#define pii pair<int, int>inline ll read() {	ll s = 0, w = 1;	char c = getchar();	while (c < '0' || c > '9') {if (c == '-') w = -1; c = getchar();}	while (c >= '0' && c <= '9') s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();	return s * w;}const int N = 100010;int n, m, a[N];struct SegmentTree {	ll sum[N << 2], lazy[N << 2];	int l[N << 2], r[N << 2];	void update(int rt) {sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];	}	void pushdown(int rt) {if (!lazy[rt]) return ;sum[rt << 1] += (r[rt << 1] - l[rt << 1] + 1) * lazy[rt], lazy[rt << 1] += lazy[rt];sum[rt << 1 | 1] += (r[rt << 1 | 1] - l[rt << 1 | 1] + 1) * lazy[rt], lazy[rt << 1 | 1] += lazy[rt];lazy[rt] = 0;update(rt);	}	void build(int rt, int L, int R) {l[rt] = L, r[rt] = R;if (L == R) {sum[rt] = a[L];return ;}int mid = L + R >> 1;build(rt << 1, L, mid), build(rt << 1 | 1, mid + 1, R);update(rt);	}	void change(int rt, int L, int R, int x) {if (L <= l[rt] && r[rt] <= R) {sum[rt] += (r[rt] - l[rt] + 1) * x;lazy[rt] += x;return ;}pushdown(rt);if (L <= r[rt << 1]) change(rt << 1, L, R, x);if (l[rt << 1 | 1] <= R) change(rt << 1 | 1, L, R, x);update(rt);	}	ll query(int rt, int L, int R) {if (L <= l[rt] && r[rt] <= R) return sum[rt];pushdown(rt);ll res = 0;if (L <= r[rt << 1]) res += query(rt << 1, L, R);if (l[rt << 1 | 1] <= R) res += query(rt << 1 | 1, L, R);return res;	}} tree;int main() {	n = read(), m = read();	for (int i = 1; i <= n; i++) a[i] = read();	tree.build(1, 1, n);	while (m--) {int op = read();if (op == 1) {int l = read(), r = read(), x = read();tree.change(1, l, r, x);}if (op == 2) {int l = read(), r = read();printf("%lld\n", tree.query(1, l, r));}	}	return 0;}

现在，你学会了区间修改区间查询的线段树，不如想想单点修改区间查询和区间修改单点查询怎么写？（为什么没有单点修改单点查询线段树呢（小声）
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