浙江大学PAT上机题解析之1007. Maximum Subsequence Sum (25)

Given a sequence of K integers { N₁, N₂, ..., N_K }.  A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } where 1 <= i <= j <= K.  TheMaximum 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

//之前这道题是很简单的动态规划，但是提交了很多次一直有一个case通不过，后来加了条件判断，终于AC了，貌似通不过的CASE是 -1 0 -2  正确结果应该是0 0 0

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int val[10010];

int main()
{

	

	int K;
	bool flag=true;
	int sum=0;
	int max=-1;
	int temp=0;
	int first=10010;
	int end=0;
	int len=0;
	scanf("%d",&K);
		memset(val,0,sizeof(val));
		int i=0,j=0;

		while(K--)
		{
			scanf("%d",&val[i]);
			if (val[i]>=0) flag = false;
			i++;
		}

		len = end = i-1;


		for (i=0;val[i]!='';i++)
		{
			sum+=val[i];
			if (sum>max)
			{
				max = sum;
				end = i;
				first = temp;
			}

			else 
				if (sum<0)
				{
					sum = 0;
					temp = i+1;
				}

		}


		if (flag)
		{
			max = 0;
			first = 0;
			end = len;
		}
		else if (max<0)//此处为额外判断
		{
          for (i=0;i<=len;i++)
			  if (val[i]==0)
			  {
				  max = 0;
				  first = i;
				  end = i;
				  break;
			  }

		}
		printf("%d %d %d
",max,val[first],val[end]);
	

	//system("pause");
	return 0;
}