Hdu-5875 Function(树上倍增st算法)

本文链接：https://blog.youkuaiyun.com/u014258433/article/details/52496803

本文介绍了一种基于倍增技巧解决区间查询问题的方法。针对给定数组与多个查询请求，利用预处理技术和数据结构优化，实现高效计算指定区间的特殊函数值。文章通过实例详细解析了算法的设计思路及实现过程。

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Problem Description

The shorter, the simpler. With this problem, you should be convinced of this truth.

You are given an array

A of

N postive integers, and

M queries in the form

(l,r) . A function

F(l,r) (1≤l≤r≤N) is defined as:

F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate

F(l,r) , for each query

(l,r) .

Input

There are multiple test cases.

The first line of input contains a integer

T , indicating number of test cases, and

T test cases follow.

For each test case, the first line contains an integer

N(1≤N≤100000) .
The second line contains

N space-separated positive integers:

A1,…,AN (0≤Ai≤109) .
The third line contains an integer

M denoting the number of queries.
The following

M lines each contain two integers

l,r (1≤l≤r≤N) , representing a query.

Output

For each query

(l,r) , output

F(l,r) on one line.

Sample Input

Sample Output

题意：给n个数字的数列，m个询问，问F[l,r]是多少，F[l,r] = F[l,r-1] % a[r]，当l = r时F[l,r] = a[l].

分析：注意每个数最多取有效模log(a[i])次就会变成0，所以我们可以把数列中每个元素向它右边第一个小于等于它的元素连边，最后会形成一个森林，每次询问在线在森林上倍增。n*log(n)*log(a[i]).

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x3f3f3f3f
#define N 100005
using namespace std; 
typedef long long ll;
int T,n,m,fa[N][35],a[N];
int Find_pos(int u,int val)
{
	int now = log2(n-u+1) + 1;
	while(now >= 0)
	{
		if(a[fa[u][now]] > val && a[fa[u][now+1]] <= val) u = fa[u][now];
		now--;
	}
	if(a[u] > val) u = fa[u][0];
	return u;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		memset(fa,0,sizeof(fa));
		a[0] = -1;
		scanf("%d",&n);
		for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
		for(int i = n-1;i;i--)
		{
			fa[i][0] = Find_pos(i+1,a[i]);
			for(int j = 1;(1<<j) <= n;j++)
			 fa[i][j] = fa[fa[i][j-1]][j-1];
		}
		scanf("%d",&m);
		for(int i = 1;i <= m;i++)
		{
			int l,r,ans;
			scanf("%d%d",&l,&r);
			int now = l+1;
			ans = a[l];
			while(now && now <= r && ans)
			{
				ans %= a[now];
				now = Find_pos(now,ans);
			}
			printf("%d\n",ans);
		}
	}
}