HDUOJ----1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 113031    Accepted Submission(s): 26130

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

 

Author

Ignatius.L

http://acm.hdu.edu.cn/showproblem.php?pid=1003

动态规划....求最大子段和；

代吗：

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    int t,n,i,k,st,en,maxn,sum,flag=1;
    cin>>t;
    while(t--)
    {
        cin>>n;
        vector<int>arr(n);
        for(i=0;i<n;i++)
            cin>>arr[i];
          st=en=sum=0;
        for(i=0,k=0;i<n;i++)
         {
            sum+=arr[i];
            if(i==0||sum>maxn)
           {
                maxn=sum;
                en=i;
                st=k;
           } 
          if(sum<0)
           {
             sum=0;
             k=i+1;
           }
         }
         cout<<"Case "<<flag++<<":"<<endl;
         cout<<maxn<<" "<<st+1<<" "<<en+1<<endl;
         if(t!=0)cout<<endl;
    }
    return 0;
}