hdu4027 Can you answer these queries?

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这个题乍一看没思路啊，但后来发现每个数小于2^63，开方至多6次就变成1了，然后就不用开方了，只有这么暴力了。
（似乎我的代码里有一些没用的东西= =）

#include<algorithm>#include<cstdio>#include<cmath>using namespace std;typedef long long LL;const int MAXN = 100000;int n;LL A[MAXN + 10];struct seg {    int l, r;    LL sum;    inline int length()    {        return r - l + 1;    }    void update(seg &lch, seg &rch)   {        if(l < r)  sum = lch.sum + rch.sum;    }}   Tree[MAXN * 4 + 10];void Build(int u, int l, int r)    {    seg &cur = Tree[u];    cur.l = l, cur.r = r;    if(l >= r)  cur.sum = A[l];    else    {        int mid = (l + r) / 2;        Build(u * 2, l, mid);        Build(u * 2 + 1, mid + 1, r);        cur.update(Tree[u * 2], Tree[u * 2 + 1]);    }}void Ins(int u, int l, int r)  {    seg &cur = Tree[u], &lch = Tree[u * 2], &rch = Tree[u * 2 + 1];    if(cur.r < l || cur.l > r)  return ;    else {        if(l <= cur.l && cur.r <= r && cur.length() == cur.sum)    return ;        else if(cur.l >= cur.r)  cur.sum = sqrt(cur.sum + 0.5);        else Ins(u * 2, l, r), Ins(u * 2 + 1, l, r);    }    cur.update(lch, rch);}LL Query(int u, int l, int r) {    seg &cur = Tree[u];    if(cur.r < l || cur.l > r)  return 0;    else if(l <= cur.l && cur.r <= r)   return cur.sum;    else return Query(u * 2, l, r) + Query(u * 2 + 1, l, r);}int main()  {    int kase = 0;    while(scanf("%d", &n) == 1 && n)    {        for(int i = 1; i <= n; i++)            scanf("%I64d", &A[i]);        Build(1, 1, n);        int ops, opt, l, r;        scanf("%d", &ops);        printf("Case #%d:\n", ++kase);        for(int i = 1; i <= ops; i++)   {            scanf("%d%d%d", &opt, &l, &r);            if(l > r)   swap(l, r);            if(opt) printf("%I64d\n", Query(1, l, r));            else Ins(1, l, r);        }        puts("");    }    return 0;}

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