数据结构基础 - Maximum Subsequence Sum(子序列基础上修改)

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本文探讨了最大子序列和问题的求解方法，通过一个具体示例介绍了如何找到具有最大元素之和的连续子序列及其起始和结束位置。针对特殊情况进行了详细说明，并提供了完整的C语言代码实现。

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01-复杂度2 Maximum Subsequence Sum（25 分）
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

代码如下，代码的解释很详细了，
因为这道题有天杀的7个测试点，爆炸！！

#include <stdio.h>
#include <stdlib.h>
int K;
int a[100100];
int first,last,len;
int main(){
    scanf("%d",&K);
    int MaxSum = -1;   /*Zero could also be the MaxSum---a special case here*/
    int ThisSum;
    for(int i=0;i<K;i++){
        scanf("%d",&a[i]);
    }
    ThisSum=0;
    first=0;
    last=K-1;
    int tmpFirst=first;
    for(int i=0;i<K;i++){
        ThisSum=ThisSum+a[i];
        if(ThisSum > MaxSum){
            MaxSum=ThisSum;
            last=i;     /*Remember the last one*/
        }
        if(ThisSum < 0){
            tmpFirst=i+1;/*Remember the first one, which is the next*/
            ThisSum=0;
        }
    }
    if(MaxSum < 0){
    //All the inegrates are negative, so we need to set the first to zero And set the MaxSum to zero because MaxSum is -1 now*/
        tmpFirst=first;
        MaxSum=0;
    }else{
    /*If MaxSum is positive, we should check is the subsequence is unique or not?
     and output the one with the smallest indices i and j
     I used a stupid algorithm because i did the for cycle again
     But given that the O(N), i think it's acceptable*/
        ThisSum=0;      /*Don't forget initialize it again*/
        for(int i=0;i<K;i++){
            ThisSum=ThisSum+a[i];
            if(ThisSum == MaxSum){
                last=i;  /*Remember the last one*/
                break;
            }
            if(ThisSum < 0){
                ThisSum=0;
                tmpFirst=i+1;
            }
        }
    }
    printf("%d %d %d",MaxSum,a[tmpFirst],a[last]);
}