HDU OJ 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 255908    Accepted Submission(s): 60826

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

 

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code：

#include<iostream>  
using namespace std;  
#define Min -999999  
int main()  
{  
    int data[100000],start,end;  
    int m;  
    int step=1;  
    cin>>m;      
    while(m--)  
    {  
        int n;  
        cin>>n;  
        for (int i=1; i<=n;i++)  
            cin>>data[i];  
        int max = Min;  
        int k=1;  
        int sum = 0;  
        for (int i=1; i<=n; i++)  
        {  
            sum = sum + data[i];  
            if (sum > max)  
            {  
                max = sum;  
                start=k;  
                end=i;  
            }  
            if(sum<0)  
            {  
                sum=0;  
                k=i+1;  
            }  
        }  
        if(step!=1)  
            cout<<endl;  
        cout<<"Case "<<step<<":"<<endl;  
        cout<<max<<" "<<start<<" "<<end<<endl;  
        step++;  
    }  
    return 0;  
}  

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