The Water Problem


Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 425    Accepted Submission(s): 339



Problem Description


a1,a2,a3,...,anrepresenting 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


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)


 



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


 



Source


2015 ACM/ICPC Asia Regional Changchun Online 






#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

using namespace std;

const int N = 20010;
int maxx[N][30];
int minn[N][30];
int n,m;

void play(){
    int l = floor(log10(double(n))/log10(double(2)));
    for(int j=1;j<=l;j++){
        for(int i=1;i<=n+1-(1<<j);i++){
            maxx[i][j] = max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
        }
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int x,y;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            maxx[i][0] = x;
            minn[i][0] = x;
        }
        play();
        scanf("%d",&m);
        while(m--){
            scanf("%d%d",&x,&y);
            int pp = floor(log10(double(y-x+1))/log10(double(2)));
            printf("%d\n",max(maxx[x][pp],maxx[y-(1<<pp)+1][pp]));

        }
    }
    return 0;
}