bzoj3211.花神游历各国(线段树)

            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: JDD
PROG: bzoj3211.花神游历各国 
DATE: 2015.11.04
*/

#include 
#include  

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;
}