点击链接PAT甲级-AC全解汇总

题目：
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
1
2

Sample Output:

10 1 4
1

题意：
找出一列数中，和最大的连续的子列
注意： 输出的是和 第一个数 最后一个数，而不是第一个数和最后一个数的下标。此外，还要注意case5 存在输入没有正数但是有0，其他都是负数的情况。

我的代码：

#include
using namespace std;

int main()
{
    int N,sum=0,left_flag=0;
    int res_sum=-1,left=0,right=0;
    cin>>N;
    int num[N]={0};
    bool flag_non_pos=true;
    for(int i=0;i<N;i++)
    {
        cin>>num[i];
        sum+=num[i];
        if(sum>res_sum)
        {
            res_sum=sum;
            left=num[left_flag];
            right=num[i];
            flag_non_pos=false;
        }
        else if(sum<0)
        {
            sum=0;
            left_flag=i+1;
        }
    }
    if(flag_non_pos)
        printf("0 %d %d",num[0],num[N-1]);
    else
        printf("%d %d %d",res_sum,left,right);

    return 0;
}

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