ZOJ 3714 Java Beans

Java Beans

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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

累加java beans的值，注意kids是围成圈的。

#include "stdio.h"
using namespace std;

int jb[205];
int sum[500];
int main()
{
    int ca, n, m, a, max;
    scanf("%d", &ca);
    while (ca--){
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++)
            scanf("%d", &jb[i]);
        sum[0] = jb[0];
        for (int i = 1; i < n + m - 1; i++){
            sum[i] = sum[i - 1]+jb[i%n];
        }
        max = sum[m-1];
        for (int i = m; i < n + m - 1; i++){
            if (sum[i] - sum[i - m]>max) max = sum[i] - sum[i - m];
        }
        printf("%d
", max);
    }
}