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;
}