HDU 1003 Max sum (maximum sum of consecutive subcolumns, Dynamic Planning)

Max sum

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 55792 accepted submission (s): 12580

Problem descriptiongiven 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.

Inputthe 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 ).

Outputfor 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 input2
5 6-1 5 4-7
7 0 6-1 1-6 7-5

Sample outputcase 1:
14 1 4

Case 2:
7 1 6

Authorignatius. L Algorithm : 1. always add until there is a negative number and a maximum sum can be obtained during the addition, record
2. Repeat Step 1 to update the largest sum until the input is complete.
Principle: If the substring <A1, A2, A3,... is in a row ,..., an> medium Sn <0; S1, S2 ,... S (n-1) are both greater than or equal to 0
Then there must be the largest Si, that is, the largest sum of the continuous substrings from A1 to.
We know that when A1> 0, the sum of any <AI... an> (I> 1) is smaller than 0,
Therefore, these elements cannot form the largest and largest substrings (if a1 = 0, A2> = 0) with an element );
At the same time, the sum of <AI... AJ> (I> 1, j <= N) cannot be greater than that of Si.

 /*  
Hdu1003
Dynamic Planning
  */   
# Include  <  Stdio. h  >  
  # Define  Maxn 100000 
  Int  Main ()
{
  Int  T, icase, I, sum, maxsum, J, X, Y,;
  Int  N;
Scanf (  "  % D  "  ,  &  T );
Icase  =  0  ;
  While (T  --  )
{
Icase  ++  ;
Sum  =  0  ;
Maxsum  =-  20000  ;
I  =  1  ;
Scanf (  "  % D  " ,  &  N );
  For  (J  =  1  ; J  <=  N; j  ++  )
{
Scanf (  "  % D  "  ,  & A );
Sum  + =  A;
  If  (Maxsum  <  Sum)
{
Maxsum  =  SUM;
X  =  I;
Y  =  J;
}
  If  (Sum  < 0  )
{
I  =  J  +  1  ;
Sum  =  0  ;
}
}
Printf (  "  Case % d: \ n  "  , Icase );
Printf (  "  % D \ n "  , Maxsum, x, y );
  If  (T  >  0  ) Printf (  "  \ N  "  );
}
  Return     0  ;
}