J - Java Beans

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本文介绍了一种解决环形序列中寻找连续子序列最大值的算法问题，通过实例演示了如何通过一次遍历实现最优解，适用于竞赛编程及算法优化场景。

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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.

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Output

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

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Sample Input

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

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Sample Output

16
158

将要查找的序列看为一个环，比较一周，找其中最大值

#include<stdio.h>
int main()
{
	int t,n,m,a[200],i,sum=0,t2,t1,T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)scanf("%d",&a[i]);
		sum=0;
		for(i=0;i<m;i++)sum+=a[i];
		if(n==m)printf("%d\n",sum);
		else{t1=0;t=sum;
			for(i=1;;i++)
			{
				if(i%n==0)break; t2=(i+m-1)%n;
				t=t-a[t1]+a[t2];
				t1=i%n;
				if(t>sum)sum=t;
				
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}