HDUOJ 4027 Can you answer these queries?

HDUOJ 4027 Can you answer these queries?

题目链接

Problem 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 263.
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

10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8

Sample Output

Case #1:
19
7
6

看似简单的线段树，实则很恶心
首先输入的区间端点不一定满足 $X<Y$，关键是线段树操作时，直接用 $min$ 和 $max$ 会 STACKOVERFLOW，这点错误我找了好久，真的无语，改正很简单，加一个 $swap$ 函数即可~
另外每组数据都要换行，此次区间操作并不是加上一个数或者全改成一个数，每个数的变化都不同，所以无法用懒惰标记，直接操作即可，AC代码如下：

#include
using namespace std;
typedef long long ll;
const int N=1e5+5;
int n,m,op,x,y,id=1;
ll tree[N<<2];
void pushup(int i)
{
     
    tree[i]=tree[i<<1]+tree[i<<1|1];
}

void build(int i,int l,int r)
{
     
    if(l==r)
    {
     
        scanf("%lld",&tree[i]);
        return ;
    }
    int mid=l+r>>1;
    build(i<<1,l,mid);
    build(i<<1|1,mid+1,r);
    pushup(i);
}

void update(int i,int l,int r,int m,int n)
{
     
    if(tree[i]==r-l+1) return;
    if(l==r){
     
        tree[i]=sqrt(tree[i]);
        return;
    }
    int mid=l+r>>1;
    if(m<=mid) update(i<<1,l,mid,m,n);
    if(n>mid) update(i<<1|1,mid+1,r,m,n);
    pushup(i);
}

ll query(int i,int l,int r,int m,int n)
{
     
    if(m<=l&&r<=n) return tree[i];
    ll ans=0;int mid=l+r>>1;
    if(m<=mid) ans=ans+query(i<<1,l,mid,x,y);
    if(n>mid) ans=ans+query(i<<1|1,mid+1,r,x,y);
    return ans;
}

int main()
{
     
    while(~scanf("%d",&n)){
     
        printf("Case #%d:\n",id++);
        build(1,1,n);
        scanf("%d",&m);
        while(m--){
     
            scanf("%d%d%d",&op,&x,&y);
            if(x>y) swap(x,y);
            if(op==0) update(1,1,n,x,y);
            else printf("%lld\n",query(1,1,n,x,y));
        }
        printf("\n");
    }
    return 0;
}