HDU 5443 The Water Problem（水）

最新推荐文章于 2020-08-23 13:49:15 发布

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最新推荐文章于 2020-08-23 13:49:15 发布

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本文介绍了一个简单的算法，用于解决寻找给定区间内的最大水源大小的问题。通过读取一系列水源大小并处理多个查询请求，该算法能够高效地找到每个查询区间内的最大水源值。

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思路：没有修改，只有询问操作，可以RMQ，线段树等各种方法，当然也可以暴力

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int a[maxn];
int main()
{
    int T;
	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 l,r,ans=-1;
			scanf("%d%d",&l,&r);
			for(int i = l;i<=r;i++)
                ans = max(ans,a[i]);
			printf("%d\n",ans);
		}
	}
}

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

representing the size of the water source. Given a set of queries each containing

integers

and

, please find out the biggest water source between

and

Input

First you are given an integer

indicating the number of test cases. For each test case, there is a number

on a line representing the number of water sources.

integers follow, respectively

, and each integer is in

. On the next line, there is a number

representing the number of queries. After that, there will be

lines with two integers

and

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

Sample Output