Hdu1231 maximum continuous subsequence (Dynamic Planning)

Maximum continuous subsequence

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 39872 accepted submission (s): 18076

The Problem description is a sequence of {N1, N2,..., nk} Given K integers. Any continuous subsequence can be represented as {Ni, Ni + 1 ,...,
NJ}, where 1 <= I <= j <= K. The maximum continuous subsequence is the element and the largest of all consecutive subsequences,
For example, for a given sequence {-2, 11,-4, 13,-5,-2}, its maximum continuous subsequence is {11,-4, 13}, the largest and
Is 20.
In this year's data structure examination paper, the maximum sum of programming requirements is required. Now, a requirement is added, that is,
The first and last elements of the subsequence.

 

The input test input contains several test cases. Each test case occupies two rows, Row 1 provides a positive integer k (<1st), and row 3 provides K integers, separated by spaces. When K is 0, the input ends and the case is not processed.

 

For each test case, output the first and last elements of the largest sum and maximum continuous subsequences in one row.
, Separated by spaces. If the maximum continuous subsequence is not unique, the smallest sequence numbers I and j are output (for example, 2nd and 3 groups in the input sample ). If all k elements are negative, the maximum value is 0, and the first and last elements of the entire sequence are output.

 

Sample Input6-2 11-4 13-5-210-10 1 2 3 4-5-23 3 7-2165-8 3 2 5 01103-1-5-23-1 0- 20

 

Sample output20 11 1310 1 410 3 510 10 100-1-20 0 0 HintHinthuge input, scanf is recommended.

Question: Give an integer sequence, find a continuous and maximum subsequence, output sum, the first element of the subsequence, And the last element of the subsequence.

Problem: state transition equation: DP [I] = max (A [I], DP [I-1] + A [I])

Emm is clearly a dynamic plan, and it always feels violent.

 

 1 #include<bits/stdc++.h> 2 using namespace std; 3 int a[10005],dp[10005]; 4 int main() 5 { 6     int k; 7     while(~scanf("%d",&k),k) 8     { 9         memset(a,0,sizeof(a));10         memset(dp,0,sizeof(dp));11         for(int i=1;i<=k;i++)12         {13             scanf("%d",&a[i]);14         }15         int maxx=-9999,s,e=-9999;16         for(int i=1;i<=k;i++)17         {18             dp[i]=max(a[i],dp[i-1]+a[i]);19             if(maxx<dp[i])20             {21                 maxx=dp[i]; 22             }    23         }24         for(int i=1;i<=k;i++)25         {26             if(dp[i]==maxx&&e==-9999)27             {28                 e=i;break;29             }30         }31         for(int i=e;i>=1;i--)32         {33             if(dp[i]==a[i])34             {35                 s=i;break;36             }37         }38         if(maxx<0)39         {40             printf("0 %d %d\n",a[1],a[k]);41         }else 42         {43             printf("%d %d %d\n",maxx,a[s],a[e]);44         }45     }46     return 0;47 }

 

Hdu1231 maximum continuous subsequence (Dynamic Planning)