HDOJ 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 85953    Accepted Submission(s): 19912

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

 1 #include <iostream>
 2 using namespace std;  3 
 4 
 5 int main()  6 {  7         int numbers;  8         cin>>numbers;  9         for(int k =0;k<numbers;k++) 10  { 11                 int num; 12                 cin>>num; 13                 
14                 
15                         int * number = new int [num]; 16                         int * start =  new int [num]; 17                         int * b =  new int [num]; 18                         for(int i=0;i<num;i++) 19                                 cin>>number[i]; 20                         
21                         b[0]= number[0]; 22                         start[0]=0; 23                         for(int i=1;i<num;i++) 24  { 25                                 if(b[i-1]+number[i]>=number[i]) 26  { 27                                         b[i]=b[i-1]+number[i]; 28                                         start[i]= start[i-1]; 29  } 30                                 else
31  { 32                                         b[i]= number[i]; 33                                         start[i]= i; 34  } 35  } 36                         int max = b[0]; 37                         int pos=0; 38                         for(int i=0;i<num;i++) 39  { 40                                 if(b[i]>max) 41  { 42                                         max=b[i]; 43                                         pos = i; 44  } 45  } 46                         cout<<"Case "<<k+1<<":"<<endl<< max<<" "<<start[pos]+1<<" "<<pos+1<<endl; 47                         if(k!=numbers-1) 48                                 cout<<endl; 49  delete number; 50  delete start; 51  delete b; 52 
53 
54  } 55 
56 
57         return 0; 58 
59 
60 
61 }