const int MAXN = 500000;
struct SegTree
{
int sum[MAXN * 4 + 5];
// 根据子节点算当前节点
void up(int now)
{
sum[now] = sum[now * 2] + sum[now * 2 + 1];
}
// 基于 a 数组 build
void build(int a[], int now, int l, int r)
{
if (l == r)
{
sum[now] = a[l];
return;
}
int mid = (l + r) / 2;
build(a, now * 2, l, mid);
build(a, now * 2 + 1, mid + 1, r);
up(now);
}
// 当前节点是 now,对应区间是 l~r
// 希望给第 x 个数加上 y
void update(int now, int l, int r, int x, int y)
{
if (l == r)
{
sum[now] += y;
return;
}
int mid = (l + r) / 2;
if (x <= mid)
update(now * 2, l, mid, x, y);
if (x > mid)
update(now * 2 + 1, mid + 1, r, x, y);
up(now);
}
// 当前节点是 now,对应区间是 l~r
// 返回 x~y 在这个区间内的和
int query(int now, int l, int r, int x, int y)
{
if (x <= l && r <= y)
return sum[now];
int mid = (l + r) / 2;
int res = 0;
if (x <= mid)
res += query(now * 2, l, mid, x, y);
if (y >= mid + 1)
res += query(now * 2 + 1, mid + 1, r, x, y);
return res;
}
};
const int MAXN = 100000;
struct SegTree
{
int sum[MAXN * 4 + 5];
int lazy[MAXN * 4 + 5];
// 根据子节点算当前节点
void up(int now)
{
sum[now] = sum[now * 2] + sum[now * 2 + 1];
}
// 把当前节点的lazy传下去
void down(int now, int l, int r)
{
if (lazy[now] == 0)
return;
int mid = (l + r) / 2;
sum[now * 2] += (mid - l + 1) * lazy[now];
sum[now * 2 + 1] += (r - mid) * lazy[now];
lazy[now * 2] += lazy[now];
lazy[now * 2 + 1] += lazy[now];
lazy[now] = 0;
}
// 基于 a 数组 build
void build(int now, int l, int r)
{
if (l == r)
{
sum[now] = a[l];
return;
}
int mid = (l + r) / 2;
build(now * 2, l, mid);
build(now * 2 + 1, mid + 1, r);
up(now);
}
// 当前节点是 now,对应区间是 l~r
// 希望给第 x~y 个数加上 z
void update(int now, int l, int r, int x, int y, int z)
{
if (x <= l && r <= y)
{
lazy[now] += z;
sum[now] += (r - l + 1) * z;
return;
}
int mid = (l + r) / 2;
down(now, l, r);
if (x <= mid)
update(now * 2, l, mid, x, y, z);
if (y >= mid + 1)
update(now * 2 + 1, mid + 1, r, x, y, z);
up(now);
}
// 当前节点是 now,对应区间是 l~r
// 返回 x~y 在这个区间内的和
int query(int now, int l, int r, int x, int y)
{
if (x <= l && r <= y)
return sum[now];
int mid = (l + r) / 2;
int res = 0;
down(now, l, r);
if (x <= mid)
res += query(now * 2, l, mid, x, y);
if (y >= mid + 1)
res += query(now * 2 + 1, mid + 1, r, x, y);
return res;
}
};