poj2593

Max Sequence

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 13533		Accepted: 5654

Description

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

You should 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 20

0

Sample Output

40

Source

POJ Monthly--2005.08.28,Li Haoyuan

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 

 5 using namespace std;

 6 

 7 int dp1[100010],dp2[100010];

 8 int a[100010];

 9 int n;

10 int maxD;

11 

12 int main()

13 {

14 

15     while(scanf("%d",&n)!=EOF && n)

16     {

17         memset(dp1,0,sizeof(dp1));

18         memset(dp2,0,sizeof(dp2));

19         for(int i=1; i<=n; i++)

20         {

21             scanf("%d",&a[i]);

22         }

23         //dp for 最大子序列和

24         dp1[0]=-10000000;

25         for(int i=1; i<=n; i++)

26         {

27             if(dp1[i-1]<0)

28                 dp1[i]=a[i];

29             else

30                 dp1[i]=dp1[i-1]+a[i];

31         }

32         for(int i=1; i<=n; i++)

33         {

34             if(dp1[i-1]>dp1[i])

35             {

36                 dp1[i]=dp1[i-1];

37             }

38         }

39 

40         //reverse dp

41         dp2[n+1]=-10000000;

42         for(int i=n; i>=1; i--)

43         {

44             if(dp2[i+1]<0)

45                 dp2[i]=a[i];

46             else

47                 dp2[i]=dp2[i+1]+a[i];

48         }

49         for(int i=n; i>=1; i--)

50         {

51             if(dp2[i+1]>dp2[i])

52                 dp2[i]=dp2[i+1];

53         }

54 

55         //cal

56         maxD=-10000000;

57         for(int i=1; i<=n-1; i++)

58         {

59             if(dp1[i]+dp2[i+1] > maxD)

60                 maxD=dp1[i]+dp2[i+1];

61         }

62 

63         printf("%d\n",maxD);

64     }

65     return 0;

66 }