bzoj3211.花神游历各国(线段树)

来源：互联网发布：中级工程师java面试题编辑：程序博客网时间：2024/06/10 08:53

一个序列，有两个操作：1、将一段区间的数开方； 2、求一段区间的和；

因为开平方不支持区间修改，只能进行单点修改，这样的话复杂度就会变成 O(n＾2)；如何优化？因为是开平方，所以10 ^ 9 的数最多进行6 - 7次左右就会变成 1 或 0，就没有必要在开方了，这里就打一个标记就可以了。不知道网上为什么那么多人要写并查集，可能会更快，还没想到。

/*Author: JDDPROG: bzoj3211.花神游历各国 DATE: 2015.11.04*/#include <cstdio>#include <cmath> using namespace std;const int MAX_N = 100005;struct node{int l, r;long long sum;bool flag;}a[MAX_N << 2];int n, m, data[MAX_N];inline int read(){int ret = 0; char c = getchar();while(!(c >= '0' && c <= '9')) c = getchar();while(c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();return ret;}void updata(int i){int t1 = i << 1, t2 = t1 + 1;a[i].sum = a[t1].sum + a[t2].sum;a[i].flag = a[t1].flag & a[t2].flag;}void make_tree(int i, int l, int r){a[i].l = l; a[i].r = r;if(a[i].l == a[i].r){a[i].sum = (long long)data[l]; if(a[i].sum == 1 || a[i].sum == 0) a[i].flag = 1;return;}int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1;make_tree(t1, l, mid); make_tree(t2, mid + 1, r);updata(i); }void tree_sqrt(int i, int l, int r){if(a[i].flag) return;if(a[i].l == a[i].r){a[i].sum = (long long)sqrt(a[i].sum);if(a[i].sum == 1 || a[i].sum == 0) a[i].flag = 1;return;}int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1;if(r <= mid) tree_sqrt(t1, l, r);else if(l > mid) tree_sqrt(t2, l, r);else tree_sqrt(t1, l, mid), tree_sqrt(t2, mid + 1, r);updata(i);}long long tree_calc(int i, int l, int r){if(l > r) return 0;if(a[i].l == l && a[i].r == r) return a[i].sum;int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1;if(r <= mid) return tree_calc(t1, l, r);else if(l > mid) return tree_calc(t2, l, r);else return tree_calc(t1, l, mid) + tree_calc(t2, mid + 1, r);}void init(){n = read();for(int i = 1; i <= n; i ++) data[i] = read();make_tree(1, 1, n);}void doit(){m = read();while(m --){int k, x, y; k = read(); x = read(); y = read();if(k == 1) printf("%lld\n", tree_calc(1, x, y));else tree_sqrt(1, x, y);}}int main(){init(); doit();return 0;}

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