P3372 线段树 1

考点

  • 线段树

题解

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 50;
int n, m, a[maxn];

struct {
int l, r;
ll sum, tag;
#define l(x) tr[x].l
#define r(x) tr[x].r
#define sum(x) tr[x].sum
#define tag(x) tr[x].tag
} tr[4 * maxn];

void build(int p, int l, int r) {
l(p) = l, r(p) = r;
if (l == r) {
sum(p) = a[l];
return;
}
int mid = (l + r) / 2;
build(p * 2, l, mid), build(p * 2 + 1, mid + 1, r);
sum(p) = sum(p * 2) + sum(p * 2 + 1);
}

// 向下传递标签
void down(int p) {
if (tag(p)) {
sum(p * 2) += tag(p) * (r(p * 2) - l(p * 2) + 1);
sum(p * 2 + 1) += tag(p) * (r(p * 2 + 1) - l(p * 2 + 1) + 1);
tag(p * 2) += tag(p), tag(p * 2 + 1) += tag(p);
tag(p) = 0;
}
}

void update(int p, int l, int r, int k) {
if (l <= l(p) && r(p) <= r) {
sum(p) += (ll)k * (r(p) - l(p) + 1);
tag(p) += k;
return;
}
down(p);
int mid = (l(p) + r(p)) / 2;
if (l <= mid) update(p * 2, l, r, k);
if (r > mid) update(p * 2 + 1, l, r, k);
sum(p) = sum(p * 2) + sum(p * 2 + 1);
}

ll ask(int p, int l, int r) {
if (l <= l(p) && r(p) <= r) return sum(p);
down(p);
int mid = (l(p) + r(p)) / 2;
ll ans = 0;
if (l <= mid) ans += ask(p * 2, l, r);
if (r > mid) ans += ask(p * 2 + 1, l, r);
return ans;
}

int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
build(1, 1, n);
int opt, x, y, k;
while (m--) {
scanf("%d", &opt);
if (opt == 1) {
scanf("%d%d%d", &x, &y, &k);
update(1, x, y, k);
} else {
scanf("%d%d", &x, &y);
printf("%lld\n", ask(1, x, y));
}
}
return 0;
}

思路

线段树的模板题,不再赘述。