【PTA】01-复杂度2 Maximum Subsequence Sum

最新推荐文章于 2020-10-14 10:39:56 发布

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最新推荐文章于 2020-10-14 10:39:56 发布

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分类专栏： PTA#数据结构

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本文详细解析了求解最大子序列和问题的两种算法，包括复杂度为O(N^2)的暴力枚举方法和更为高效的算法。通过具体实例说明了如何找到连续子序列中元素和最大的子序列，并输出其首尾元素及最大和。适用于初学者理解最大子序列和问题及其解决策略。

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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

思路：

利用课上老师讲过的算法2，4分别实现，尤其要注意当数据中只有负数和0的时候。自己处理的时候弄得比较繁琐，有用了一个函数来判断数组中是否只有负数和0。网上参考之后，将maxsum初始化为-1，这样就可以了。

代码：

算法2

#include <stdio.h>

void MaxSubseqSum2( int a[], int N);

int main(int argc,const char *argv[])
{
	int k;
	scanf("%d",&k);
	int a[k];
	for(int i=0;i<k;i++)
	{
		scanf("%d",&a[i]);
	}
	MaxSubseqSum2(a,sizeof(a)/sizeof(a[0]));
	return 0;
}

void MaxSubseqSum2( int a[], int N)
{ 
    int flag=0;
    int *p1=a,*q1=&a[N-1];//定义指针用于指向最大子列和的左端和右端 
    int thissum=0,maxsum=-1;
    for(int i=0;i<N;i++)
	{
    	thissum=0;
    	for(int j=i;j<N;j++)
		{
    		thissum+=a[j];
    		if(thissum>maxsum)
			{
				flag=1; 
				maxsum=thissum;
				p1=&a[i];
				q1=&a[j];
			}
		}
	}
    if（！flag)
   {
      printf("%d %d %d",0,*p1,*q1);
   }else
   {
      printf("%d %d %d",maxsum,*p1,*q1);
   }
}

算法4

void MaxSubseqSum4(int a[], int N)
{ 
    int *ml=a,*mr=&a[N-1];
    int *l=a;
    int thissum=0,maxsum=-1;
    int flag=0;
    for(int i=0;i<N;i++)
	{
    	thissum+=a[i];
    	if(thissum>maxsum)
		{
			flag=1;
    		maxsum=thissum;
    		ml=l;//需要注意的点
    		mr=&a[i];
		}else if(thissum<0)
		{
			l=&a[i+1];
			thissum=0;
		}
	}
	if(!flag)
	{
		printf("%d %d %d",0,*ml,*mr);
	}else
    {
	    printf("%d %d %d",maxsum,*ml,*mr);
    }
}