ZCMU训练赛-J（循环节+字符串处理）

J - Java Beans

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 Mconsecutively 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 Nintegers 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

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
int a[200];
bool cmp(const int &a,const int &b)
{
    return a>b;
}
int main()
{
    int t;
    int n,m,i,j,k;
    int a[205];
    scanf("%d",&t);
    while(t--)
    {
        int b[205]={0};
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);

        for(i=0;i<n;i++)
        {
            for(j=i,k=0;j<n+1,k<m;j++,k++)
            {
                if(j==n)
                    j=0;
                    b[i]+=a[j];
            }
        }
        sort(b,b+n);
       printf("%d
",b[n-1]);
    }
    return 0;
}

#include<stdio.h>
int d[210];
int main(int argc, char* argv[])
{
    int t,i,j;
    int n,m,sum,max;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d%d",&n,&m);
            for(i=0;i<n;i++)
            scanf("%d",&d[i]);
            
            max=0;
            for(i=0;i<n;i++)
            {
                sum=0;//注意复位
                for(j=i;j<i+m;j++)
                sum+=d[j%n];
                if(max<sum)
                    max=sum;
            }
            printf("%d
",max);
        }
    }
    return 0;
}