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HDU 4027 Can you answer these queries? (线段树)

2024-07-19 21:23:35   230人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=4027

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
 


题意：

给出一排敌军的血量，每次攻击都能将范围内的敌军血量变为原来血量的算术平方根（下取整），并询问范围内敌军的血量和。

分析：

显然的线段树，但是似乎不太好设计lazy标记啊，我们想一想算术平方根，sqrt(1)=1，且64位整数范围内最多6次就变到1了，那么只要区间内的数都为1,我就不用更新这个区间了，所以每次更新都更新到叶子结点，维护区间和就行了。数据里没有0，不过X>Y这种trick有意思吗？


/*
 *
 *	Author	:	fcbruce
 *
 *	Date	:	2014-08-27 16:46:52 
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#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/文档/code/t","r",stdin);
	#endif // ONLINE_JUDGE
	
	int n,m,__=0;
	while (~scanf( "%d",&n))
	{
		printf( "Case #%d:\n",++__);
		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 "\n",query(a,b,0,0,n));
			else
				update(a,b,0,0,n);
		}
		putchar( '\n');
	}
	
	return 0;
}


HDU 4027 Can you answer these queries? (线段树)


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