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HDU 4027 Can you answer these queries? (線段樹)

編輯：C++入門知識

HDU 4027 Can you answer these queries? (線段樹)

Can you answer these queries?
Total Submission(s): 8458 Accepted Submission(s): 1929
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)

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

Source The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

題意：

給出一排敵軍的血量，每次攻擊都能將范圍內的敵軍血量變為原來血量的算術平方根（下取整），並詢問范圍內敵軍的血量和。

分析：

顯然的線段樹，但是似乎不太好設計lazy標記啊，我們想一想算術平方根，sqrt(1)=1，且64位整數范圍內最多6次就變到1了，那麼只要區間內的數都為1,我就不用更新這個區間了，所以每次更新都更新到葉子結點，維護區間和就行了。數據裡沒有0，不過X>Y這種trick有意思嗎？

/*
 *
 *	Author	:	fcbruce
 *
 *	Date	:	2014-08-27 16:46:52 
 *
 */
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
	#define lld %I64d
#else
	#define lld %lld
#endif

#define maxm 
#define maxn 100007

using namespace std;

LL sum[maxn<<2];

inline void pushup(itn k)
{
	sum[k]=sum[k*2+1]+sum[k*2+2];
}

void build(int k,int l,int r)
{
	if (r-l==1)
	{
		scanf( lld ,&sum[k]);
		return ;
	}
	
	build(k*2+1,l,l+r>>1);
	build(k*2+2,l+r>>1,r);
	
	pushup(k);
}

void update(int a,int b,int k,int l,int r)
{
	if (b<=l || r<=a)	return ;
	if (sum[k]==r-l)	return ;
	
	if (r-l==1)
	{
		sum[k]=(LL)sqrt(sum[k]);
		return ;
	}
	
	update(a,b,k*2+1,l,l+r>>1);
	update(a,b,k*2+2,l+r>>1,r);
	
	pushup(k);
	
}

LL query(itn a,int b,itn k,int l,int r)
{
	if (b<=l || r<=a)	return 0;
	if (a<=l && r<=b)	return sum[k];
	
	return query(a,b,k*2+1,l,l+r>>1)+query(a,b,k*2+2,l+r>>1,r);
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen(/home/fcbruce/文檔/code/t,r,stdin);
	#endif // ONLINE_JUDGE
	
	int n,m,__=0;
	while (~scanf( %d,&n))
	{
		printf( Case #%d:
,++__);
		build(0,0,n);
		itn q,a,b;
		scanf( %d,&m);
		while (m--)
		{
			scanf( %d%d%d,&q,&a,&b);
			if (a>b)	swap(a,b);
			a--;
			if (q)
				printf( lld 
,query(a,b,0,0,n));
			else
				update(a,b,0,0,n);
		}
		putchar( '
');
	}
	
	return 0;
}