3001

3001

Problem A

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 260   Accepted Submission(s) : 51

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.<br>

 

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).<br>

 

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.<br>

 

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

   
   
   
   

 

简单题意：

对于给出的一个序列，求它的子序列的最大和；即求最大子序列的问题；

思路：对于总体的最大和，可以想成是求局部的最大和；

         对于每一个数，可以认为是一个独立的最大和或者是把它当成前面已经求得的最大和的增量；

         求得第一个最大和并且保存起来，然后依次求出局部最大和，就可以得到总体的最大和；

AC代码：

#include <iostream>
using namespace std;
int get(int data[] , int &l , int &r , int dl)
{
     int max = -10000000 ;
     l = 0 ;
     r = 0 ;
     int t = 1 ;
     int mt = 0 ;
     for(int i = 0 ; i < dl ; i++)
     {
             mt = mt +data[i];
             if(mt > max)
             {
                            max =  mt;
                            l = t ; r = i+1 ;
             }


             if(mt < 0)
             {
                              mt = 0 ;
                              t = i+2 ;
             }
     }
     return max ;
}
int main()
{
    int num ;
    cin >> num ;
    for(int k = 1 ; k <= num ;k++)
    {
              int lg = 0 ;
              cin >>lg ;
              int * data = new int[lg];
              for(int i =0 ;i < lg ; i++)
              {
                      cin >>data[i];
              }
              int l = 0 , r = lg-1 ;
              int max = 0 ;
              max = get(data , l , r , lg);
              cout<<"Case "<<k<<":"<<endl;
              cout<<max<<" "<<l<<" "<<r<<endl;


              if(k!=num)
              cout<<endl;
    }
    return 0 ;
}