PAT1007. Maximum Subsequence Sum

Given a sequence of K integers { N₁, N₂, ..., N_K }. A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4思路：动态规划简单题，此题错误的原因是没有仔细看题，一定要仔细的分析英文的题目。仔细分析输出的是什么。

 1 #include<cstdio>
 2 const int MAXX=10010;
 3 int data[MAXX];
 4 int dp[MAXX];
 5 int pre[MAXX];
 6 int main(int argc, char *argv[])
 7 {
 8     int count;
 9     scanf("%d",&count);
10     int flag=0;
11     for(int i=0;i<count;i++)
12     {
13         scanf("%d",&data[i]);
14         if(data[i]<0)
15            flag++;
16     }
17     if(flag==count)
18       printf("0 %d %d\n",data[0],data[count-1]);
19     else
20     {
21         dp[0]=data[0];
22         pre[0]=0;
23         for(int i=1;i<count;i++)
24         {
25             if(dp[i-1]+data[i]>=data[i])
26             {
27                 dp[i]=dp[i-1]+data[i];
28                 pre[i]=pre[i-1];
29             }
30             else
31             {
32                 dp[i]=data[i];
33                 pre[i]=i;
34             }
35
36         }
37        int index=0;
38        for(int i=0;i<count;i++)
39        {
40             if(dp[i]>dp[index])
41             {
42                  index=i;
43             }
44
45         }
46           printf("%d %d %d\n",dp[index],data[pre[index]],data[index]);
47     }
48
49
50
51     return 0;
52 }