A - The Water Problem

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.

Input

First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an, and each integer is in {1,...,106}. On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

3

1

100

1

1 1

5

1 2 3 4 5

5

1 2

1 3

2 4

3 4

3 5

3

1 999999 1

4

1 1

1 2

2 3

3 3

Sample Output

100

2

3

4

4

5

1

999999

999999

1

求区间内最大的水源，之前区间dp没法做，会T，用rmq

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include <iomanip>

#include<cmath>

#include<float.h>

#include<string.h>

#include<algorithm>

#define sf scanf

#define pf printf

#define scf(x) scanf("%d",&x)

#define scff(x,y) scanf("%d%d",&x,&y)

#define prf(x) printf("%d\n",x)

#define mm(x,b) memset((x),(b),sizeof(x))

#include<vector>

#include<queue>

//#include<map>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=a;i>=n;i--)

typedef long long ll;

const ll mod=1e9+7;

const double eps=1e-8;

const int inf=0x3f3f3f3f;

using namespace std;

const double pi=acos(-1.0);

const int N=1e3+10;

int a[N],MAX[N][20];

void first(int n)

{

	for(int j=1;(1<<j)<=n;j++)

	{

		for(int i=1;i+(1<<j)-1<=n;i++)

		{

			MAX[i][j]=max(MAX[i][j-1],MAX[i+(1<<(j-1))][j-1]);

			//MIN[i][j]=min(MIN[i][j-1],MIN[i+(1<<(j-1))][j-1];

		}

	}

}

int solve(int l,int r)

{

	int x=0;

	while(l-1+(1<<x)<=r) x++;

	x--;

	return max(MAX[l][x],MAX[r-(1<<x)+1][x]);

}

int main()

{

	int re,n,tot,l,r;scf(re);

	while(re--)

	{

		mm(a,0);

		mm(MAX,0);

		scf(n);

		rep(i,1,n+1)

		{

			scf(a[i]);

			MAX[i][0]=a[i];

		}

		first(n);

		scf(tot);

		while(tot--)

		{

			scff(l,r);

			prf(solve(l,r));

		}

	}

	return 0;

}

巴特西