hdu 4027 Can you answer these queries?(线段树)

题目大意：hdu 4027 Can you answer these queries?

题目大意：给定一个长度为N的序列，Q次操作，0 l r：将区间l r之间的数开根；1 l r：查询l r之间数的和。

解题思路：这题看上去是一道线段树，其实它就是一道线段树，只不过不用想的太复杂，因为开根的趋近1的速度非常快，所以每个节点只要标记区间内元素是否相同即可。复杂度妥妥的。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef  long long ll;

const int maxn = 100000;

int N, Q;
ll P[maxn + 5];

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2];
ll s[maxn << 2], v[maxn << 2];

inline void maintain(int u, ll d) {
v[u] = d;
s[u] = d * (rc[u] - lc[u] + 1);
}

inline void pushup(int u) {
s[u] = s[lson(u)] + s[rson(u)];
v[u] = (v[lson(u)] == v[rson(u)] ? v[lson(u)] : 0);
}

inline void pushdown(int u) {
if (v[u]) {
maintain(lson(u), v[u]);
maintain(rson(u), v[u]);
v[u] = 0;
}
}

void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;

if (l == r) {
v[u] = s[u] = P[l];
return;
}

int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid + 1, r);
pushup(u);
}

void modify(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r && v[u]) {
maintain(u, (ll)sqrt((double)v[u]));
return;
}

pushdown(u);
int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
modify(lson(u), l, r);
if (r > mid)
modify(rson(u), l, r);
pushup(u);
}

ll query(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return s[u];

pushdown(u);
ll ret = 0;
int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
ret += query(lson(u), l, r);
if (r > mid)
ret += query(rson(u), l, r);
pushup(u);
return ret;
}

int main () {
int kcas = 1;
while (scanf("%d", &N) == 1) {
for (int i = 1; i <= N; i++)
scanf("%I64d", &P[i]);
build(1, 1, N);
printf("Case #%d:\n", kcas++);

int t, l, r;
scanf("%d", &Q);
while (Q--) {
scanf("%d%d%d", &t, &l, &r);
if (l > r)
swap(l, r);
if (t)
printf("%I64d\n", query(1, l, r));
else
modify(1, l, r);
}
printf("\n");
}
return 0;
}