hdu5443 The Water Problem（长春网络赛）

The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 207 Accepted Submission(s): 166

Problem Description

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
   
   
   
   

Source

2015 ACM/ICPC Asia Regional Changchun Online 

题意：给出n个水池的水量，找出区间内最大的水量。

分析：咋一看觉得是线段树，后面一看数据范围，才10*1000*100嘛，果断暴力。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int a[1005];
int n,q,l,r;

int main ()
{
    int T;
    scanf ("%d",&T);
    while (T--)
    {
        scanf ("%d",&n);
        for (int i=1; i<=n; i++)
            scanf ("%d",&a[i]);
        scanf ("%d",&q);
        int maxx;
        while (q--)
        {
            maxx=0;
            scanf ("%d%d",&l,&r);
            for (int i=l; i<=r; i++)
                maxx=max(maxx, a[i]);
            cout<<maxx<<endl;
        }
    }
}