1007 Maximum Subsequence Sum (25分) Python解法

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本文详细解析了求解最大子序列和问题的算法，通过动态规划思想，介绍了如何找到给定整数序列中具有最大元素和的连续子序列。文章提供了具体的输入输出规格，并附带了一个示例输入和输出，帮助读者理解算法实现。

Given a sequence of K integers { N1, N2 , …, NK}. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj} 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

利用动态规划思想dp[i]=max(nums[i],nums[i]+dp[i-1])，同时记录好初始位置p和连续数组长度dis即可，q=p+dis。

a=int(input())
nums=input().split()
co=0
for i in range(a):
        nums[i]=int(nums[i])
        if nums[i]<0:
                co+=1
if(co==a):
        print(0,nums[0],nums[a-1])
else:
        n=len(nums)
        dp=[0]*n
        t=-1000
        dis=0
        d=0
        for i in range(0,n):
                if i==0 :
                        dp[0]=nums[i]
                        p=0
                        q=0
                else:
                        dp[i]=max(nums[i],nums[i]+dp[i-1])
                        if(dp[i]==nums[i]):
                                d=i
                                dis=0
                        else:
                                dis+=1
                if dp[i]>t:
                        t=dp[i]
                        p=d
                        q=dis+d
        print(t,nums[p],nums[q])