P5459 回转寿司

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;
const ll lx = -1e10 - 1e9, rx = 1e10;
ll L, R, a[maxn];
int n, tot, root;

struct node {
ll lc, rc;
ll cnt;
#define lc(x) tr[x].lc
#define rc(x) tr[x].rc
#define cnt(x) tr[x].cnt
} tr[55 * maxn];

int build() {
++tot;
lc(tot) = rc(tot) = cnt(tot) = 0;
return tot;
}

void up(int p) { cnt(p) = cnt(lc(p)) + cnt(rc(p)); }

void update(int p, ll l, ll r, ll x) {
if (l == r) {
cnt(p) += 1;
return;
}
ll mid = l + (r - l) / 2;
if (x <= mid) {
if (!lc(p)) lc(p) = build();
update(lc(p), l, mid, x);
}
if (x > mid) {
if (!rc(p)) rc(p) = build();
update(rc(p), mid + 1, r, x);
}
up(p);
}

ll ask(int p, ll l, ll r, ll L, ll R) {
if (L <= l && r <= R) return cnt(p);
ll mid = l + (r - l) / 2, ans = 0;
if (L <= mid && lc(p)) ans += ask(lc(p), l, mid, L, R);
if (R > mid && rc(p)) ans += ask(rc(p), mid + 1, r, L, R);
return ans;
}

int main() {
root = build();
update(root, lx, rx, 0);
cin >> n >> L >> R;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
a[i] += a[i - 1];
}
ll ans = 0;
for (int i = 1; i <= n; ++i) {
ans += ask(root, lx, rx, a[i] - R, a[i] - L);
update(root, lx, rx, a[i]);
}
cout << ans;
return 0;
}

思路

对原数组求前缀和保存到数组a,那么区间和就简化为L <= a[r] - a[l - 1] <= R

移两边,式子变成a[r] - R <= a[l - 1] <= a[r] - L

从左到右遍历前缀和数组,每个点a[i]当做a[r],求区间[a[r] - R, a[r] - L]有多少个点。

(当然,要手动把0加到前缀和数组里,0也算左端点a[l - 1]

使用权值线段树,由于这里数据范围太大,最小值是-1e5 * 1e5 - 1e9,最大值是1e5 * 1e5 - 1e9

采用动态开点线段树即可