题目链接：点击打开链接

题意：

给定n长的序列

以下2个操作

0 x y 给[x,y]区间每一个数都 sqrt

1 x y 问[x, y] 区间和

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <math.h>
#include <vector>
#include <set>
using namespace std;
#define ll long long
#define N 100005
#define L(x) tree[x].l
#define R(x) tree[x].r
#define Lson(x) (x<<1)
#define Rson(x) (x<<1|1)
#define Num(x) tree[x].num
#define Sum(x) tree[x].sum
inline int Mid(int x, int y){return (x+y)>>1;}
struct Subtree{int l, r;ll num, sum;
}tree[N<<2];
ll a[N];
void push_up(int id){if(L(id) == R(id))return;Sum(id) = Sum(Lson(id)) + Sum(Rson(id));if(Num(Lson(id)) <= 1  && Num(Rson(id)) <= 1)Num(id) = 1;else Num(id) = 2;
}
void build(int l, int r, int id){L(id) = l; R(id) = r;if(l == r) { Num(id) = Sum(id) = a[l]; return ;}int mid = Mid(l, r);build(l, mid, Lson(id));build(mid+1, r, Rson(id));push_up(id);
}
void updata(int l, int r, int id){if(L(id) == R(id)){Sum(id) = Num(id) = (ll)sqrt((double)Sum(id));return ;}if(l == L(id) && R(id) == r && Num(id) <= 1) return ;int mid = Mid(L(id), R(id));if(mid < l)updata(l, r, Rson(id));else if(r <= mid)updata(l, r, Lson(id));else {updata(l, mid, Lson(id));updata(mid+1, r, Rson(id));}push_up(id);
}
ll query(int l, int r, int id){if(l == L(id) && R(id) == r)return Sum(id);int mid = Mid(L(id), R(id));if(mid < l)return query(l, r, Rson(id));else if(r <= mid)return query(l, r, Lson(id));else return query(l, mid, Lson(id)) + query(mid+1, r, Rson(id));
}
int n;
void solve(){int q, op, x, y;for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);build(1, n, 1);scanf("%d",&q);while(q--){scanf("%d %d %d", &op, &x, &y);if(x > y) swap(x, y);if(op == 0)updata(x, y, 1);elseprintf("%lld\n", query(x, y, 1));}
}
int main(){int Cas = 1;while(~scanf("%d", &n)){printf("Case #%d:\n", Cas++);solve();puts("");}return 0;
}