J

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 Ci (1 ≤ Ci ≤ 1000) indicating number of java beans the ith 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

题意：

题意为N个小朋友围成一圈，每个小朋友手里都有一些豆子，老师想抢连续的M个小朋友的豆子，问老师最多能抢多少？

思路：

运用for循环，因为是一个圆形，所以最后一个完了之后了以在从第一个开始（如果第一个没被取）！

细节：

用sort排序之后，输出最后一个即可！

代码：

#include<bits/stdc++.h> using namespace std;int main(){int t,n,m,i,j;cin>>t;while(t--){int p[402];int pp[201]={0};int max;cin>>n>>m;for(i=0;i<n;i++){cin>>p[i];p[i+n]=p[i];}for(i=0;i<n;i++){int sum=0;max=0;for(j=i;j<i+m;j++){pp[i]+=p[j];}}sort(pp,pp+n);cout<<pp[n-1]<<endl;}return 0;}

不够细心！！！！