hdu 5443 The Water Problem（长春网络赛——暴力）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5443

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1084    Accepted Submission(s): 863

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,...,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

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

 

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题目大意：给出n个水池的水量，找出区间内最大的水量。

求区间最值。正常都採用线段树的方法。可是这题数据量不大。全部暴力就过了~

详见代码。

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int t;
    int a[1010],l,r;
    scanf("%d",&t);
    while (t--)
    {
        int n;
        scanf("%d",&n);
        for (int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        int q;
        scanf("%d",&q);
        while (q--)
        {
            int Max=0;
            scanf("%d%d",&l,&r);
            for (int i=l; i<=r; i++)
            {
                if (a[i]>Max)
                    Max=a[i];
            }
            cout<<Max<<endl;
        }
    }
    return 0;
}