Java Beans ZOJ - 3714_javabeans zoj-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41524782/article/details/80787032

本文介绍了一种解决最大连续子数组求和问题的方法，该问题是寻找数组中长度为M的最大连续子数组的总和。文章提供了一个有效的算法实现，并通过示例说明了输入输出的格式。

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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

AC代码（sum的初值要放对位置）

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t, n, m, a[1000+10], b[1000+10], sum, maxi, i, j;
    cin>>t;
    while(t--)
    {
     sum = 0, maxi = 0;
     cin>>n>>m;
     for(i = 0;i<n;i++)
     {
      cin>>a[i];
      b[i] = a[i];
     }
     if(n==m)
     {
      for(i = 0;i<n;i++)
      {
       sum = sum + a[i];
      }
      cout<<sum<<endl;
     }
     else
     {
      for(i = 0;i<n;i++)
      {
       a[i+n] = b[i];
      }
      for(i = 0;i<=2*n-m;i++)
      {
        sum = 0;
       for(j = 0;j<m;j++)
       {
        sum = sum + a[i+j];
       }
       if(maxi<sum)
       {
        maxi = sum;
       }
      }
       cout<<maxi<<endl;
      }
     }

    return 0;
}