Can you answer these queries?+线段树

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 6423    Accepted Submission(s): 1483

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
 
/*思路：基本线段树+lazy思想
因为大部分的数经多次开根号后，都会变成1，所以设置一个flag标志，
如果为true就不再往下面更新了*/
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define MAX 100000
typedef struct infor
{
	int  l , r ;
	bool flag;
	__int64 val ;
}infor ;
infor t[4*MAX] ;
__int64 a[MAX+1] ;
void create(int rt , int l , int r )
{
	t[rt].l = l ;
	t[rt].r = r ;
	t[rt].flag =  false ;
	if(l == r)
	{
		t[rt].val = a[l] ;
		if(a[l]==1)
			t[rt].flag = true ;
		return ;
	}
	int mid = (l + r) / 2 ;
	create(rt*2 , l , mid );
	create(rt*2 + 1 , mid + 1 , r);
	t[rt].flag = t[rt*2].flag && t[rt*2 + 1].flag  ;
	t[rt].val = t[rt*2].val + t[rt*2 + 1].val ;

}
void update(int rt , int l , int r , int x , int y)
{
	if(t[rt].flag )
		return ;
	if(t[rt].l == t[rt].r ){
		t[rt].val = (__int64)(sqrt((double)t[rt].val));
		if(t[rt].val <= 1)	
			t[rt].flag = true ;
		return ;
	}
	int mid = (t[rt].l + t[rt].r ) / 2 ;
	if(x > mid )
		update(rt*2 + 1 , mid + 1 , r , x, y);
	else if(y<=mid )
		update(rt*2 , l , mid , x , y );
	else
	{
		update(rt*2 , l , mid , x , y);
		update(rt*2 + 1 , mid + 1 , r , x , y );
	}
	t[rt].flag = t[rt*2].flag && t[rt*2 + 1].flag ;
	t[rt].val = t[rt*2].val + t[rt*2 + 1 ].val ;
}
__int64 cal(int rt , int l , int r , int x , int y )
{
	if(x <= t[rt].l && y>=t[rt].r )
		return t[rt].val ;
	int mid = (t[rt].l + t[rt].r ) / 2 ;
	if(x>mid )
		return cal(rt*2 + 1 , mid + 1 , r , x , y );
	else if(y<= mid )
		return cal (rt*2 , l , mid , x , y ) ;
	else 
	{
		return cal(rt*2 , l , mid , x , y ) + cal (rt*2 + 1 , mid + 1 , r , x  , y ) ;
	}
}
int main( void )
{
	int n ,count = 1 ;
	while(scanf("%d",&n)!=EOF)
	{
		int i ;
		for( i = 1 ; i <= n ; i ++)
			scanf("%I64d",&a[i]);
		create(1 , 1 , n );
		int t ;
		printf("Case #%d:\n",count++);
		scanf("%d",&t);
		while(t--)
		{
			int q , x , y ;
			scanf("%d %d %d",&q,&x,&y);
			if(x>y)
			{
				int temp;
				temp = x ; 
				x = y ;
				y = temp ;
			}
			if(q==0)
				update(1 , 1 , n , x , y );
			else
				printf("%I64d\n",cal(1 , 1 , n , x , y));
		}
		printf("\n");
	}
	return 0;
}