J. Java Beans

最新推荐文章于 2021-06-15 00:19:21 发布

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本文介绍了一个关于JavaBeans的游戏挑战问题，旨在寻找连续座位上孩子持有的JavaBeans最大总和。通过多组测试案例，利用循环和累积求和的方法解决此问题，并提供了一段C++实现代码。

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J. Java Beans

Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

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There are N little kids sitting in a circle, each of them are carrying some java beans in their hand. Their teacher want to select M kids who seated in M consecutive seats and collect java beans from them.

The teacher knows the number of java beans each kids has, now she wants to know the maximum number of java beans she can get from M consecutively seated kids. Can you help her?

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases.

For each test case, the first line contains two integers N (1 ≤ N ≤ 200) and M (1 ≤ M ≤ N). Here N and M are defined in above description. The second line of each test case contains N integers C_i (1 ≤ C_i ≤ 1000) indicating number of java beans the i^th kid have.

Output

For each test case, output the corresponding maximum java beans the teacher can collect.

Sample Input

2
5 2
7 3 1 3 9
6 6
13 28 12 10 20 75

Sample Output

16
158

坑点 ：一个环

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
int a[2000];
int main()
{
	//std::ios::sync_with_stdio(false);
	int t;
	while (cin>>t)
	{
		while (t--)
		{
			int n, m;
			int i;
			cin >> n >> m;
			for (i = 0; i < n; i++)
			{
				cin >> a[i];
				a[n + i] = a[i];
			}
			
			//sort(a, a + n);
			long long sum = 0;
			int MAX = -888888;
			int  j;
			for (i = 0; i < n; i++)
			{
				sum = 0;
				for (j = i;j < i + m; j++)
					sum += a[j];
				if (sum > MAX)
					MAX = sum;
			}
				
			printf("%d\n", MAX);
		}
	}
	return 0;
}