Java BeansZOJ Problem Set - 3714 the 10th 浙江ACM赛
2013-05-12 15:56
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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 selectM 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 fromM consecutively seated kids. Can you help her?
For each test case, the first line contains two integers N (1 ≤ N ≤ 200) andM (1 ≤
M ≤ N). Here N and M are defined in above description. The second line of each test case containsN integers
Ci (1 ≤ Ci ≤ 1000) indicating number of java beans theith kid have.
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 selectM 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 fromM 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) andM (1 ≤
M ≤ N). Here N and M are defined in above description. The second line of each test case containsN integers
Ci (1 ≤ Ci ≤ 1000) indicating number of java beans theith 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<stdio.h> int main() { int T,N,M,i,j; int arr[440]; int max=-1,sum; scanf("%d",&T); while(T--) {sum=0;max=-1; scanf("%d%d",&N,&M); for(i=0;i<N;i++) { scanf("%d",&arr[i]); arr[i+N]=arr[i]; } for(i=0;i<N;i++) {sum=0; for(j=i;j<i+M;j++) sum+=arr[j]; if(sum>max) max=sum; } printf("%d\n",max); } return 0; }
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