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

HDU 4027 Can you answer these queries?

题目链接

题意：给定一个数列。两种操作

0 a b 把[a,b]区间内的数字都开根

1 a b 询问区间[a,b]和

思路：注意开根最多开到1或0就不在变化，那么一个数字最多开63次，然后题目保证数列和小于2^63，所以实际上对于每一个数字的改动总次数并不多，因此改动操作每次就单点改动，线段树多开一个标记，表示这个区间是否所有都已经不变了

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 100005;

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
int l, r;
ll sum;
bool cover;
} node[4 * N];

int n;

void pushup(int x) {
node[x].cover = (node[lson(x)].cover && node[rson(x)].cover);
node[x].sum = node[lson(x)].sum + node[rson(x)].sum;
}

void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r; node[x].cover = false;
if (l == r) {
scanf("%I64d", &node[x].sum);
if (node[x].sum == 0 || node[x].sum == 1) node[x].cover = true;
return;
}
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
pushup(x);
}

void add(int l, int r, int x = 0) {
if (node[x].cover) return;
if (node[x].l == node[x].r) {
node[x].sum = (ll)sqrt(node[x].sum * 1.0);
if (node[x].sum == 1) node[x].cover = true;
return;
}
int mid = (node[x].l + node[x].r) / 2;
if (l <= mid) add(l, r, lson(x));
if (r > mid) add(l, r, rson(x));
pushup(x);
}

ll query(int l, int r, int x = 0) {
if (node[x].l >= l && node[x].r <= r)
return node[x].sum;
int mid = (node[x].l + node[x].r) / 2;
ll ans = 0;
if (l <= mid) ans += query(l, r, lson(x));
if (r > mid) ans += query(l, r, rson(x));
return ans;
}

int main() {
int cas = 0;
while (~scanf("%d", &n)) {
build(1, n);
scanf("%d", &n);
int op, a, b;
printf("Case #%d:\n", ++cas);
while (n--) {
scanf("%d%d%d", &op, &a, &b);
if (a > b) swap(a, b);
if (op == 0) add(a, b);
else printf("%I64d\n", query(a, b));
}
printf("\n");
}
return 0;
}