这道题看一眼就知道是线段树的题目，但是一看，题目要求区间开方，我就又傻了。想了一会，发现就算是109在开方5次之后就变成1了，所以就算是我们单个开方，时间复杂度也就是O(NlogN)。但是为了避免重复开方，我们要给线段树的每一个节点做一个标记，表示下面的区间是否全部为1和0。那样我们就可以避免重复开方了。

代码：

#include 
     
      #include 
      
       #define LL long longinline void swap(int &a, int &b) {
    int c = a; a = b; b = c;}#define MAXN 100005inline void GET(LL &n){    n = 0; char c;    do c = getchar(); while(c > '9' || c < '0');    while(c <= '9' && c >= '0') {n = n * 10 + c - '0'; c = getchar();}}inline void GET(int &n){    n = 0; char c;    do c = getchar(); while(c > '9' || c < '0');    while(c <= '9' && c >= '0') {n = n * 10 + c - '0'; c = getchar();}}struct node{    LL sum;    bool lazy;    node(){lazy = 0;}}t[MAXN<<1];inline int idx(int l, int r) {
    return (l + r) | (l != r);}void pushup(int l, int r){    int mid = (l + r) >> 1;    t[idx(l, r)].sum = t[idx(l, mid)].sum + t[idx(mid+1, r)].sum;    t[idx(l, r)].lazy = t[idx(l, mid)].lazy && t[idx(mid+1, r)].lazy;}void build(int l, int r){    int i = idx(l, r);    if(l == r)    {        GET(t[i].sum);        return;    }    int mid = (l + r) >> 1;    build(l, mid); build(mid+1, r);    t[idx(l, r)].sum = t[idx(l, mid)].sum + t[idx(mid+1, r)].sum;}int op, L, R, n, m;void Ins(int l, int r){    int i = idx(l, r);    if(l > R || r < L) return;    if(t[i].lazy) return;    if(l == r)    {        t[i].sum = sqrt(t[i].sum);        if(t[i].sum == 1 || t[i].sum == 0) t[i].lazy = 1;        return;    }    int mid = (l + r) >> 1;    Ins(l, mid);    Ins(mid+1, r);    pushup(l, r);}LL Sum(int l, int r){    if(l > R || r < L) return 0;    if(l >= L && r <= R) return t[idx(l, r)].sum;    int mid = (l+r)>>1;    return Sum(l, mid) + Sum(mid+1, r);}int main(){    GET(n);    build(1, n);    GET(m);    while(m --)    {        GET(op);GET(L);GET(R);        if(op == 2)            Ins(1, n);        else if(op == 1)            printf("%lld\n", Sum(1, n));    }    return 0;}