Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

SP2713 GSS4 - Can you answer these queries IV

LuoGu: SP2713 GSS4 - Can you answer these queries IV

SPOJ: GSS4 - Can you answer these queries IV

考虑如何处理区间开方的操作。每个数最大为 $10 ^ {18}$ ,这意味着最多开方 $\log_2 18 + 1$ 次后,这个数为 $1$ ,此时无论如何开方,值不会变。所以,维护一个区间最大值,当区间最大值为 $1$ 时,区间内的所有数都是 $1$ ,不需要再操作。

查看代码
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#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int n, m;
LL w[N];
struct Node
{
int l, r;
LL mx, sum;
}tr[N << 2];
void pushup (int x)
{
tr[x].mx = max(tr[x << 1].mx, tr[x << 1 | 1].mx);
tr[x].sum = tr[x << 1].sum + tr[x << 1 | 1].sum;
}
void build (int x, int l, int r)
{
tr[x].l = l;
tr[x].r = r;
if (l == r)
{
tr[x].mx = tr[x].sum = w[l];
return;
}
int mid = l + r >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
pushup(x);
}
LL query (int x, int l, int r)
{
if (tr[x].l >= l && tr[x].r <= r)
return tr[x].sum;
int mid = tr[x].l + tr[x].r >> 1;
LL res = 0;
if (l <= mid)
res += query(x << 1, l, r);
if (r > mid)
res += query(x << 1 | 1, l, r);
return res;
}
void dfs (int x, int l, int r)
{
if (tr[x].mx <= 1)
return;
if (tr[x].l == tr[x].r)
{
tr[x].mx = tr[x].sum = sqrt(tr[x].mx);
return;
}
int mid = tr[x].l + tr[x].r >> 1;
if (l <= mid)
dfs(x << 1, l, r);
if (r > mid)
dfs(x << 1 | 1, l, r);
pushup(x);
}
int main ()
{
int cas = 1;
while (~scanf("%d", &n))
{
printf("Case #%d:\n", cas++);
memset(tr, 0, sizeof tr);
for (int i = 1; i <= n; i++)
scanf("%lld", &w[i]);
build(1, 1, n);
scanf("%d", &m);
for (int k, l, r; m; m--)
{
scanf("%d%d%d", &k, &l, &r);
if (l > r)
swap(l, r);
if (k)
printf("%lld\n", query(1, l, r));
else
dfs(1, l, r);
}
puts("");
}
return 0;
}