Max Sum

最新推荐文章于 2019-07-31 16:31:45 发布

李逝水

最新推荐文章于 2019-07-31 16:31:45 发布

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本文讨论了MaxSum问题的求解方法，并通过两个示例解释了如何找到序列中最大子序列的和及其起始和结束位置。程序采用贪心算法策略，通过迭代计算并更新最大和及其对应的子序列范围。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 99625 Accepted Submission(s): 22922

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

#include<iostream>
using namespace std;
int main()
{
int t,count=1;
cin>>t;
while(t--){

  int n,a[100002],max=-100000,begin,end,sum=0,k=0;
  cin>>n;
  for(int i=0;i<n;i++){
   cin>>a[i];
   sum+=a[i];
   if(sum>max)
   {
    max=sum;
    end=i+1;
    begin=k+1;
   }
   if(sum<0)
   {
    k=i+1;
    sum=0;
   }
  }

  cout<<"Case "<<count++<<":"<<endl;
  cout<<max<<" "<<begin<<" "<<end<<endl;
  if(t!=0)
   cout<<endl;

}
return 0;
}

难点：位置的确定