POJ 2593 & amp; 2479: Max Sequence

POJ 2593 & 2479: Max Sequence
Max Sequence

Time Limit:3000 MS		Memory Limit:65536 K
Total Submissions:16329		Accepted:6848

Description

Give you N integers a1, a2... aN (| ai |<= 1000, 1 <= I <= N ).

 

You shoshould output S.

 

Input

The input will consist of several test cases. for each test case, one integer N (2 <= N <= 100000) is given in the first line. second line contains N integers. the input is terminated by a single line with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5-5 9 -5 11 200

Sample Output

40

The question is to give a sequence and find the maximum value of the sum of the two subsequences in the sequence. I switched to POJ2479 two or three years ago, but I still didn't understand dp at the time. (Of course, the degree of understanding of dp can also be switched to dp questions ...). So I feel infinite emotion when doing this question. In fact, it is very simple to calculate the sum of a sequence, that is, dp [I] = max (dp [I-1] + value [I], value [I]) now it requires the maximum sum of the two sequences. So I want to come from the left and from the right. Left [I] indicates the maximum sequence and on the left of the number from 1st to the current number I. Right [I] indicates the maximum sequence sum on the right from the number Test (from right to left) to the number I.
Code:

# Include
 
  
# Include
  
   
# Include using namespace std; int left_v [100005]; int right_v [100005]; int value [100005]; int main () {int Test; while (cin> Test) {if (! Test) break; left_v [0] = 0; right_v [0] = 0; left_v [Test + 1] = 0; right_v [Test + 1] = 0; int I, max_v =-100000000; for (I = 1; I <= Test; I ++) {cin> value [I];} left_v [1] = value [1]; right_v [Test] = value [Test]; for (I = 2; I <= Test; I ++) {left_v [I] = max (left_v [I-1] + value [I], value [I]) ;}for (I = Test-1; I> = 1; I --) {right_v [I] = max (right_v [I + 1] + value [I], value [I]);} for (I = 2; I <= Test; I ++) {left_v [I] = max (left_v [I-1], left_v [I]);} for (I = Test-1; i> = 1; I --) {right_v [I] = max (right_v [I + 1], right_v [I]);} for (I = 1; I
   
    
Max_v) max_v = left_v [I] + right_v [I + 1];} cout <
    
     
I am very happy to drop this question. .