P4145 花神游历各国

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 = 5e5 + 50;
int n, m;
ll a[maxn];

struct {
int l, r;
ll mx, sum;
#define l(x) tr[x].l
#define r(x) tr[x].r
#define mx(x) tr[x].mx
#define sum(x) tr[x].sum
#define lc(x) x * 2
#define rc(x) x * 2 + 1
} tr[4 * maxn];

void up(int p) {
sum(p) = sum(lc(p)) + sum(rc(p));
mx(p) = max(mx(lc(p)), mx(rc(p)));
}

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

void update(int p, int l, int r) {
if (l <= l(p) && r(p) <= r && l(p) == r(p)) {
mx(p) = sqrt(mx(p));
sum(p) = sqrt(sum(p));
return;
}
int mid = (l(p) + r(p)) / 2;
if (l <= mid && mx(lc(p)) > 1) update(lc(p), l, r);
if (r > mid && mx(rc(p)) > 1) update(rc(p), l, r);
up(p);
}

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

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

思路

开方操作并不能直接对求和操作进行修改。

但根据题意,每个数字最大只有1012,只需要开方6次就能变成1;

就算每个数字都暴力开方至1,总共的时间复杂度也不过6*logn*n,小的可怜

所以设置一个区间最大值,假设该区间的最大值等于1,相当于整个区间都为1,就没有开方的必要。