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

Author: FAN, Yuzhe
Contest: The 10th Zhejiang Provincial Collegiate Programming Contest

题目大意：有N个人坐成一圈，每个人有Ci个糖果，老师想找M个连续坐的同学中获得最多的糖果，问最多几个？

解题思路：最大连续和问题，这里连续数字个数为M,采用b[i]维护前i个糖果总和，那么求从i+1开始M个的总和就是：b[i+M]-b[i],枚举从i=0到i=N-1求最大值即可。

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int main(){
 5     int T;
 6     cin>>T;
 7     int a[205],b[405];
 8     while(T--){
 9         int N,M;
10         cin>>N>>M;
11         for(int i=0;i<405;i++)b[i]=0;
12         for(int i=0;i<N;i++){
13             cin>>a[i];
14             if(i==0)b[i]=a[i];
15             else b[i]=b[i-1]+a[i];
16         }//边输入边维护一个前i个数之和b[i]
17         for(int i=N;i<2*N;i++){
18             b[i]=b[i-1]+a[i%N];
19         }//继续维护b[i]使之满足一个环遍历的要求
20         int max=-1;
21         for(int i=0;i<N;i++){
22             int sum=b[i+M]-b[i];
23             if(sum>max)max=sum;
24         }//取得长为M的最大连续和
25         cout<<max<<'\n';
26     }return 0;
27 
28 }

本文转自beautifulzzzz博客园博客，原文链接： http://www.cnblogs.com/zjutlitao/p/3590370.html ，如需转载请自行联系原作者