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.
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 }
