HDU 4027 Can you answer these queries?[区间求和+区间更新]

Description

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help. 
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line. 

Notice that the square root operation should be rounded down to integer.

Input

The input contains several test cases, terminated by EOF. 
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000) 
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2 ⁶³. 
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000) 
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive. 

Output

For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.

Sample Input

101 2 3 4 5 6 7 8 9 1050 1 101 1 101 1 50 5 81 4 8

Sample Output

Case #1:1976

题目大意：两个操作，0代表对区间进行开方处理，1代表对区间求和；这里因为和不会超过2^63，所以意味着开方不会太多次，但是额外注意一个优化点，就是开方1是多余的，所以这里要优化掉；但是直接等于1还是会TLE,所以写成 sum[rt]  = r - l + 1  ,就是不需要继续往下递归了；

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long long#define N  100050using namespace std;ll sum[N<<2] , n , m ;inline void pushup(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];    return ;}void build(int l , int r , int rt){    sum[rt] = 0 ;    if(l==r)    {        scanf("%lld",&sum[rt]);        return ;    }    int m = (l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int L ,int R , int l , int r , int rt){    if(l==r)    {        sum[rt] = sqrt(sum[rt]);        return ;    }    if(sum[rt]==r-l+1&&L<=l&&R>=r) return ;    int m = (l+r)>>1;    if(L<=m) update(L,R,lson);    if(R>m)  update(L,R,rson);    pushup(rt);}ll query(int L , int R , int l , int r , int rt){    if(L<=l&&R>=r) return sum[rt];    int m = (l+r)>>1;    ll ans = 0 ;    if(L<=m) ans+=query(L,R,lson);    if(R>m) ans+=query(L,R,rson);    return ans ;}int main(){    int t=0;    while(cin>>n)    {        build(1,n,1);        int a , b , c , x , y ;        cin>>m;        printf("Case #%d:\n", ++t);        while(m--)        {            scanf("%d%d%d",&a,&b,&c);            x = min(b,c);            y = max(b,c);            if(a==0) update(x,y,1,n,1);            else printf("%lld\n",query(x,y,1,n,1));        }        puts("");    }    return 0;}

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