01-复杂度2 Maximum Subsequence Sum

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分类专栏：数据结构MOOC 文章标签： MOOC

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博客围绕最大子项和问题展开，给出题目描述、输入输出规范及示例。指出该题是最大子项和的变形，强调两个测试点注意事项：全是负数时输出sum=0及首尾数；负数和0交替时输出0,0,0。还提及在线学习及算法复杂度为O(N2)。

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

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

这个题是最大子项和的变形，测试点有两个点需要注意
1.全是负数时，应该输出sum=0，第一个和最后一个数
2.负数和0交替时，应该输出0,0,0

在线学习

#include<stdio.h>
#define MAXSIZE 100000
int end=-2,start=-2,flag=0,t=0;
int findmax(int a[],int k)
{
	int max=0,sum=0;
	for(int i=0;i<k;i++)
	{
		sum+=a[i];
		if(max<sum)
		{
			max=sum;
			end=i;
            start=flag;
		}
		else if(sum<0)
		{
			sum=0;
            flag=i+1;
		}

	}
	return max;
}
void main()
{
	int K,i,a[MAXSIZE],sum=0;
	scanf("%d",&K);
	for(i=0;i<K;i++)
	{
		scanf("%d",&a[i]);
        if(a[i]>=0)
        {
            t=1;
        }
	}
	sum=findmax(a,K);
	if(!t)
	{
		printf("0 %d %d",a[0],a[K-1]);	
	}
	else 
	{
		printf("%d %d %d",sum,a[start],a[end]);
	}
   
}

两层循环，算法复杂度为O(N²)

#include<stdio.h>
#define MAXSIZE 100000
int start=-1,end=-1,t=0,flag=-1;
int findmax(int a[],int k)
{
	int max=-1,sum=0;
	for(int i=0;i<k;i++)
	{
		sum=0;
		for(int j=i;j<k;j++)
		{
			sum+=a[j];
			if(max<sum)
			{
				max=sum;
				end=j;
				start=i;
			}
			
		}

	}
	return max;
}
void main()
{
	int K,i,a[MAXSIZE],sum=0;
	scanf("%d",&K);
	for(i=0;i<K;i++)
	{
		scanf("%d",&a[i]);
        if(a[i]>=0)
        {
            t=1;
        }
	}
	sum=findmax(a,K);
	if(sum<=0 && t==0)
	{
		printf("0 %d %d",a[0],a[K-1]);	
	}
	else 
	{
		printf("%d %d %d",sum,a[start],a[end]);
	}
}