ACM刷题之ZOJ————Java Beans

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这是一个关于JavaBeans的游戏问题，目标是在N个孩子中找到M个连续的孩子，这些孩子的JavaBeans总数最大。通过输入每个孩子的JavaBeans数量，算法可以计算出教师能收集到的最大JavaBeans数。

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Java Beans

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

签到水题

下面是ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;

int a[11100];

int main()
{
	//freopen("f:/input.txt", "r", stdin);
	int zu,i,j,k,n,m,maxx,ma;
	scanf("%d",&zu);
	while(zu--)
	{
		CLR(a,0);
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			a[i+n]=a[i];
		}
		maxx=0;
		for(i=0;i<m;i++)
		{
			maxx+=a[i];
		}
		ma=maxx;
		for(i=0;i<2*n;i++)
		{
			
			ma-=a[i];
			ma+=a[m+i];
			//printf("%d\n",ma);
			if(ma>maxx)
			maxx=ma;
		}
		
		printf("%d\n",maxx);
		
	}
}