第一周编程作业: Maximum Subsequence Sum

最新推荐文章于 2022-10-27 17:29:32 发布

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最新推荐文章于 2022-10-27 17:29:32 发布

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文章标签：数据结构

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本文介绍了一种寻找整数序列中最大子序列和的算法，并通过实例详细解释了算法的工作原理。该算法能有效处理包含负数的情况，并在所有数字均为负数时给出特定的解决方案。

01-复杂度2 Maximum Subsequence Sum（25 point(s)） Given a sequence of K
integers { N 1 , N 2 , …, 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

程序实现如下：

#include <stdio.h>
#include <stdlib.h>
//void print_arr(int n, int * arr);

int main(void)
{
    int n = 0;
    scanf("%d", &n);
    int * arr = (int *)malloc(n * sizeof(int));

    for (int i = 0; i < n; i++)
    {
        scanf("%d", &arr[i]);
        getchar();
    }

    //int arr[5]={1,2,-8,2,1};
    //n=5;

    //n=10;
    //int arr[10]={10,-2,5,5,-1,-100,100,-2,1,2};
    //int arr[10]={-1,-2,-3,-4,-8,-3,-8,-9,-7,-11};//all the number are negative
    //int arr[10]={-5,-2,8,-1,9,1,-10,0,0,0};//the sum have a negative number
    //int arr[10]={};//all the number is zero
    //int arr[10]={0,0,10,-1,-1,5,-1,-1,-1,-100};
    //int arr[10]={-2,-5,1,50,-1,-1,-2,-300,-4,-8};
    //int arr[10]={100,-8,-8,1,-8,-1000,1,2,0,0};

    //int arr[10]={-1,-2,-3,-5,-1,0,-1,-1,-1,-2};

    int next_sum, current_sum;
    next_sum = current_sum = 0;

    int start=0, end=n-1;//final result
    int start1=0,end1=n-1;//maybe the final result
    int flag=0;
    int tempSum=0;
    int s=0;
    int j=0;
    int k=1;
    int t=0;

   // int flag2 = 0;
    //int start_temp=0;

    for (int i = 0; i < n; i++)//check all the numbers
    {
        current_sum += arr[i];
        if (flag == 0)
        {
            if (arr[i] > 0|| arr[i]==0)//find a posstive number
            {
                start1 = i;//mark it
                t=start1;
                //start=start1;
                //printf("**********%d\n",start1);
                if(arr[i]==0)
                {
                    k++;
                }
                flag = 1;
            }
            else//can't find a suitable number
            {
                current_sum=0;
                //printf("positive!!\n");
                k++;
                continue;

            }
        }
        if ( current_sum < 0)
        {
            s=current_sum;
            tempSum=current_sum-arr[i];//maybe the maxsum
            start1=i+1;
            j=i-1;
            while(arr[j]<0&&s==tempSum)
            {
                s=s-arr[j];
                end=j-1;
                //start=start1;
                j--;
               // printf("*******%d\n",end);
            }

            if(tempSum<current_sum)//when it's larger than before
            {

                start=start1;
                end=end1;
            }

            current_sum = 0;
            flag=0;

        }
        else if (current_sum > next_sum)
        {
            next_sum = current_sum;
            end1 = i;
           // printf("*******end1=%d\n",end1);
            if(tempSum<current_sum)//when it's larger than before
            {
                start=start1;
                end=end1;
            }
        }



    }
    if(k==n)
    {
        start=t;
        end=t;
    }

   // printf("less than max=%d\n",tempSum);
   // printf("sum=%d %d %d\n", next_sum,start,end);
//    for(int i=start;i<stop+1;i++)
//    {

//        next_sum=next_sum+arr[i];
//    }
    printf("%d %d %d", next_sum, arr[start], arr[end]);
    //print_arr(n, arr );


    return 0;
}